Concise Complex Analysis Solution of Exercise Problems Ai Shu Xue March 9, 2008
1
Contents 1 Calculus
3
2 Cauchy Integral Theorem and Cauchy Integral Formula
14
3 Theory of Series of Weierstrass
27
4 Riemann Mapping Theorem
52
5 Differential Geometry and Picard’s Theorem
53
6
53
A First Taste of Function Theory of Several Complex Variables
7 Elliptic Functions
53
8 Th The e Riem Rieman ann n ζ -Function and The Prime Number Theory
53
2
Contents 1 Calculus
3
2 Cauchy Integral Theorem and Cauchy Integral Formula
14
3 Theory of Series of Weierstrass
27
4 Riemann Mapping Theorem
52
5 Differential Geometry and Picard’s Theorem
53
6
53
A First Taste of Function Theory of Several Complex Variables
7 Elliptic Functions
53
8 Th The e Riem Rieman ann n ζ -Function and The Prime Number Theory
53
2
This is a solution solution manual of selected selected exercise exercise problems problems from Concise from Concise Complex Analysis , Second Second Edition, Edition, by Gong Sheng (World (World Scientific, Scientific, 2007). This version version solves the exercise problems problems in Chapter Chapter 1-3, except except the following: Chapter 1 problem 37-42; Chapter 2 problem 47, 49; Chapter 3 problem 15 (xi).
1
Calculus
1. Proof. See, for example, Munkres [4], 38.
§
2. (1)
| − i| = √ 2, arg(1 − i) = 74 π. |3 + 4i| = 5, arg(3 + 4i 4 i) = arctan 43 ≈ 0. 0 .9273 5 | − 5+12 5+12ii| = 13, arg(−5+12 5+12ii) = arccos − 13 ≈ 1. 1.9656 (Matlab command:
Proof. 2i = 2, arg(2i arg(2 i) = π2 . 1 (Matlab command: atan(4/ atan(4/3)). acos( 5/13)).
| |
−
(2) Proof. (1 + 3i 3i)3 =
18ii. −26 − 18
10 4 3i
8 6 2−3i − = 5 + 5 i. 4+i =
5 17
− 1417 i. (1 + i)n + (1 − i)n = 2
n 2 +1
cos n4 π .
(3)
| − 3i(2 − i)(3 + 2i2i)(1 + i)| = 3√ 130.
Proof. 3.
(4 3i)(2 i) (1+i)(1+3i)
−
−
√
= √ 52··√ 510 = 52 .
√ √
√
π
1 Proof. Let θ = arctan 15 , α = arctan 239 . Then 5 i = 26 26ee−iθ , 1 + i = 2ei 4 and (5 i) i )4 (1 + i) i ) = π − − i( π θ αi 4 ) 4 676 2e 4 . Meanwhile Meanwhile (5 i) (1 + i + i)) = 956 4i = 676 2e . So we must have 4 4θ = α + 2kπ 2 kπ,, 1 1 k Z, i.e. π4 = 4 arctan arctan 15 arctan 239 + 2kπ, kπ , k Z. Since Since 4 arctan arctan 15 arctan 239 +2π +2 π > 0 π2 +2 + 2π = 74 π > π4 1 and 4 arctan arctan 15 arctan 239 2π < 4 π2 2π = 0 < π4 , we must have k = 0. Theref Therefore ore π4 = 4 arctan arctan 15 1 arctan 239 .
√ ∈
− −
−
−
· −
∈
−
−
− −
−
−
−
−
4. Proof. If z = x + yi + yi,, 3
2
3
1 z¯
=
x x2 +y 2
+
y x2 +y2 i.
z 2 = x2
2
x +xy +x+i( y x y+y) . (x2 y 2 +1)2 +4x2 y 2
−
− −
− y 2 + 2xyi 2 xyi..
1+z 1 z
− =
1 x2 y2 (1 x)2 +y2
− − −
+
2y (1 x)2 +y2 i.
z z 2 +1
−
=
5. Proof. a = 1, b 1, b = α = α + iβ , c = γ = γ + + iδ . So ∆ = b = b 2
ac = (α2 − β 2 − 4γ ) + i(2αβ (2αβ − − 4ac = − 4δ ) and √ (α + iβ ) ± ∆ − z = . 2
6. 2
Proof. Denote arg z by θ, then z + rz = reiθ + 2 re iθ rer iθ = r( r (eiθ e−iθ ) = 2ir sin θ = 2iImz Imz .
−
r2 reiθ
−
7.
3
2
− rz
= r(eiθ + e −iθ ) = 2r cos θ = 2Rez 2Rez , and z
=
Proof. If a a = r = r 1 eiθ1 and b and b = r = r 2 eiθ2 , then a b r1 r2 ei(θ2 −θ1 ) = . 1 ab a ¯b 1 r1 r2 ei(θ2 −θ1 )
Denote θ Denote θ 2
− −
− −
by θ,, we can reduce the problem to comparing |r1 − r2 eiθ |2 and |1 − r1 r2 eiθ |2 . Note − θ1 by θ = r 12 − 2r1 r2 cos θ + r22 |r1 − r2eiθ |2 = (r1 − r2 cos θ)2 + r22 sin2 θ = r
and
|1 − r1r2eiθ |2 = (1 − r1r2 cos θ)2 + r12r22 sin2 θ = 1 − 2r1r2 cos θ + r12r22. b So |1 − r1 r2 eiθ |2 − |r1 − r2 eiθ |2 = (r12 − 1)(r 1)(r22 − 1). This observation shows 1a−−ab a ¯b = 1 if and only if at least b |a| < 1 one of a a and b has modulus 1; 1a−−ab < 1 and |b| < 1. < 1. ¯b < 1 if | a
8.
Proof. We prove by induction. The equation clearly holds for n = n = 1. Assume it holds for all n all n with n with n Then
N +1 +1
ai bi
i=1
2
2
| | | | | | | | − | − | | | | | | | || | | | | | | | | | | − − − | | | | − | − | − | − | | | | | − | − | | | | | | | − | − N
=
ai bi + aN +1 bN +1
N
=
i=1
N
+ aN +1
ai bi
2
bN +1
2
N
¯ + ¯aN +1 +1 bN +1
¯i¯bi a
ai bi + aN +1 +1 bN +1
i=1
i=1
N
=
¯ +1 a ¯i¯bi + a ¯N +1 +1 bN +1
i=1
2
N
=
N
ai bi + aN +1 bN +1
i=1
N . ≤ N .
i=1
N
ai
2
bi
i=1
i=1
+ aN +1
2
ai¯bj
2
aj ¯bi 2 + aN +1
2
bN +1 2 + bN +1 +1
2
1 i
i i
≤
≤
bi
≤
|ai|2
2
1 i
≤
ai
2
bN +1 2 + aN +1
2
bi
ai a ¯N +1 bi¯bN +1
2
a ¯i aN +1¯bi bN +1 +1
1 i
≤
N +1 +1
=
=
N +1 +1
ai
2
bi
i=1
i=1
N +1 +1
N +1 +1
ai
i=1
2
aj ¯bi
≤
2
≤
ai¯bj
aj ¯bi 2 .
1 i
≤
≤
n 2 By method of mathematical induction, = in=1 ai 2 in=1 bi i=1 ai bi for all n all n N. Cauchy’s inequality follows directly from this equation.
∈
aj ¯bi
1 i
≤
2
ai¯bj
2
1 i
bi
i=1
ai¯bj
2
2
≤ ai¯bj
1 i
≤
aj ¯bi
|2 holds
9. Proof. We first consider the special case where a 1 = 0 and a and a 2 = 1. Then for a for a 3 = re iθ , a 1 , a 2 , a 3 are vertices of an equilateral triangle if and only if r = 1 and cos θ = 12 . Me Mean anwhi while le the equati equation on 1 + re + re 2iθ = reiθ is equivalent to 1 + r cos2θ cos2θ = r = r cos θ r sin2θ sin2θ = r = r sin θ,
which implies r = 1 and cos θ = 12 . So the equality is proven for the special case a 1 = 0, a2 = 1. For general general case, note a21 + a 22 + a 23 = a1 a2 + a 2 a3 + a 3 a1 if and only if (a ( a2 a 1 )2 + (a3 2
−
− a 1)2
=
a1 a3 −a1 (a2 a1 )(a )(a3 a1 ), which is further equivalen equivalentt to 1 + aa32 − −a1 = a2 −a1 . This shows the general case can be reduced to the special case of a of a 1 = 0 and a and a 2 = 1. Essentially, these reductions correspond to the composition of coordinate translation and coordinate rotation.
−
−
4
10. Proof.
− − −
α α ¯ 1 α γ α ¯ γ γ¯ 1 α ¯ ¯ β β 1 β γ β γ γ ¯ 1 = D := = β γ γ γ¯ 1 0 0 1 Suppose α Suppose α
− γ = r1eiθ
1
α β
− γ = = r 2 eiθ . Then α ¯ − γ γ¯ r1 eiθ r1 e−iθ = ¯ − β − γ γ¯ r2eiθ r2e−iθ
and β and β
− −
γ γ
− − −
2
1
1
2
2
− −
= r 1 r2 ei(θ1 −θ2 )
γ α ¯ ¯ γ β
− γ γ¯ − − γ γ¯ .
− r1r2e−i(θ −θ ). 1
2
So D So D = = 0 if and only if θ θ 1 θ2 = kπ, kπ , which is equivalent to α, α , β , γ being γ being colinear. Note the above reduction corresponds to the coordination transformation that places γ γ at origin.
−
11. z¯ Proof. We apply formula (1.11) (correction: the formula for x 2 should be x be x 2 = (1+z − z = 1 |z|2 )i ). If z 2 2 2 4 3 12 x1 = 3 , x 2 = 3 , and x and x 3 = 3 . If z z = 4 + 3i 3i, then x then x 1 = 13 , x 2 = 13 , and x and x 3 = 13 .
−
− i, then
12. Proof. Denote by x = (x1 , x2 , x3 ) and y = (y1 , y2 , y3 ) the points on the Riemann sphere S 2 corresponding to z to z 1 and z and z 2 , respectively. Then x Then x and y are two ends points of a diagonal of S S 2 if and only if x x + y = 0. By formula (1.11), x + y = 0 is equivalent to
z1 +¯ z1 1+ z1 2 z1 z¯1 1+ z1 2 z1 2 1 z1 2 +1
z2 +¯ z2 | | + 1+|z2 |2 = 0 − + z2 −z¯22 = 0 | | 1+|z2 | | | − + | z2 |2 −1 = 0. | | |z2 |2 +1 0 .
Solving the third equation gives z1 z2 = 1. So we can can assum assumee z1 = reiα and z2 = 1r eiβ (r > 0). 0). Then Then 1 1 2 2 2 2 1 + z1 = 1 + r and 1 + z2 = 1 + r2 = r2 (1 + z1 ). So the other two equations become
| |
| |
|
|
| |
2r cos α + 2r 2r cos β = = 0 2r sin α + 2r 2r sin β = = 0,
which is equivalent to
β α −β cos α + 2 cos 2 = 0 β α −β sin α + 2 cos 2 = 0.
So we must have 12 (α
= 2kπ + π. So z So z 1 z¯2 = e i(α−β ) = e i(2kπ +π) = −1. − β ) = kπ + π2 (k ∈ Z), i.e. α − β =
13. Proof. Suppose z Suppose z C is on a circle. circle. Then there there exists z0 C and R > 0, such that z z0 = R. R . Suppose 2 2 2 z = x = x + iy and iy and z 0 = x 0 + iy0 . Then z z0 = R = R implies implies (x (x x0 ) + (y y0 ) = R . After After simplification, simplification, this 2 2 2 2 can be written as z + z¯0 z + z0 z¯ + z0 R = 0. Let A Let A = 1, B = z¯0 and C and C = z0 R2 . Then A, Then A, C R 2 and B > AC . ¯ z¯ + C = Conversely, if A A z 2 + Bz + Bz + B = 0 is the equation satisfied by z by z , we have two cases to consider: (i) A = 0; (ii) A (ii) A = 0. ¯ | B|2 = |B|2 C > 0. B ¯ z¯ + C In case (i), A z 2 + Bz + B + C = = 0 can be written as z 2 + B z + z ¯ + 0. Define Define A A A2 A2 A
| |
z0 = to z
¯ B A
∈ | | ||
||
∈ −
| − | | |−
2
∈ ∈
| | − − C A . The equation can be written as |z|2 + z¯0z + z0z¯+ |z0|2 = R2, which is equivalent
| and R and R = = |B A2 z0 = R = R.. So z So z is on a circle.
| − |
−
| − | | | −
5
¯ z + C = 0 stands for a straight line in C. Indeed, suppose B = a + ib In case (ii), the equation Bz + B¯ ¯ z + C = 0 becomes 2ax 2by + C = 0. If we regard straight lines as circles, and z = x + iy, then Bz + B¯ 2 ¯ +C ¯ = 0 represents a circle in case (ii) as well. A z + Bz + B To see the necessary and sufficient condition for the image of this circle on the Riemann sphere S 2 to be a great circle, we suppose a generic point z on this circle corresponds to a point (x1 , x2 , x3 ) on the Riemann +ix2 2 ¯ z + C = 0 as sphere S 2 . Then by formula (1.10) z = x11− x3 , we can write the equation A z + Bz + B¯ ¯ 1 + i(B ¯ B)x2 + (A C )x3 = (A + C ). (x1 , x2 , x3 ) falls on a great circle if and only if (x1 , x2 , x3 ) (B + B)x lies on the intersection of S 2 and a plane that contains (0, 0, 0), which is equivalent to A + C = 0 by the ¯ 1 + i(B ¯ B)x2 + (A C )x3 = (A + C ). equation (B + B)x
−
||
−
−
||
− −
−
−
14. Proof. Suppose Z 1 = (x1 , x2 , x3 ) and Z 2 = (y1 , y2 , y3 ). Then by formula (1.11), d(z1 , z2 )2
= (x1 y1 )2 + (x2 y2 )2 + (x3 y3 )2 = 2 2(x1 y1 + x2 y2 + x3 y3 ) z1 + z¯1 z2 + z¯2 z1 ¯ z1 z2 ¯ z2 z1 2 1 = 2 2 + + 1 + z1 2 1 + z2 2 i(1 + z1 2 ) i(1 + z2 2 ) z1 2 + 1 z1 2 + z2 2 z1 z¯2 ¯ z1 z2 = 4 (1 + z1 2 )(1 + z2 2 ) (z1 z2 )(¯z1 ¯ z2 ) = 4 . 2 (1 + z1 )(1 + z2 2 )
− − − − | | · | | | | | |− − | | | | − − | | | |
−
So d(z1 , z2 ) =
− · | |
− | |
| | − · |z2|2 − 1 | | |z2|2 + 1
z | √ (1+|2z|z| −)(1+ |z | ) . Consequently, 1
1
2
2
2
d(z1 ,
2
∞) = 2 z lim →∞ 2
− | | z1 z2
1
(1 + z1 2 ) |z12 |2 + 1
=
2 . 1 + z1 2
| |
15. Proof. (i) The distance between z and z 1 is no greater than the distance between z and z 2 . (ii) z z1 and z z2 are perpendicular to each other. (iii) The angle between z + i and z i is less than π4 . (iv) z is on an ellipse with c and c as the foci.
−
−
−
−
16. Proof. Regard C as R2 and apply Heine-Borel theorem and Bolzano-Weierstrass theorem for
Rn
17. Proof. Note max Rezn
{|
− Rez0|, |Imzn − Imz0|} ≤ |z − z0| ≤ |Rezn − Rez0| + |Imzn − Imz0|.
18. (1)
6
(n
∈ N).
Proof. We suppose a , b < . Then (zn )n and (zn )n are bounded sequences. Denote by M a common ε bound for both sequences. ε > 0, N , so that for any n > N , max zn a , zn b < M . Then for any n
| | | | ∞ ∀ ∃
{| − | | − |} − − − | | − − − | | · − − − − ≤ −
satisfying n > N 1 +
1 n
M 2 2
(define z 0 = z 0 = 0)
n
zk z
n k
k =1
−
ab
1 n
<
N
zk z
n k
k =1
N M 2
<
n
−
1 n
+
1 + n
n
k=N +1
zk zn −k
n
a)zn −k +
(zk
k=N +1
ab
a n
n
k =N +1
zn −k
− ab
n N 1
− − 1 < ε+ε+ a zk n
= 2ε + a
n
b
k=0
N n
1
n
1 N
n N 1
− −
1
zk
b .
k =0
Taking upper limits on both sides, we get
1 lim n→∞ n
Since ε is arbitrary, we conclude lim n→∞
n
k=1
1 n
zk zn −k
ab
n k=1 zk zn k
1 lim n→∞ n
−
2ε + 0 = 2ε.
ab = 0, i.e.
n
k =1
zk zn −k = ab.
(2) Proof. Apply the result of (1) with zn
≡ 1.
19.
√
Proof. (i) f (z) = z z¯ is a function of both z and z¯. So f is not differentiable. (ii) f (z) = z¯ is a function of z¯, so it is not differentiable. (iii) f (z) = Rez = 12 (z + z¯) is a function of both z and z¯. So f is not differentiable. 20. Proof. By chain rule,
∂ ∂ z¯ g(f (z)) = g
(f (z)) ∂f (z) = 0. So g (f (z)) is also holomorphic. ∂ z¯
21. (1) Proof. Let u(x, y) = Ref (z) and v(x, y) = Imf (z). Then by C-R equations f (z) = ux (x, y) + ivx (x, y) = vy (x, y) iu y (x, y). So f (z) 0 if and only if vx = vy = ux = uy 0. By results for functions of real variables, we conclude u and v are constants on D. so f is a constant function on D.
−
≡
≡
(2) Proof. (i) and (ii) are direct corollaries of C-R equations. For (iii), we assume without loss of generality that f (z) 0 on D. We note f (z) 2 = u 2 (z) + v 2 (z). So ∂ ∂ by the C-R equations, 0 = ∂x f (z) 2 = 2uux + 2vv x = 2uux 2vu y and 0 = ∂y f (z) 2 = 2uuy + 2vv y = u v ux u v 2uuy + 2vu x , i.e. = 0. Since f 0, is invertible on D1 := z D : f (z) = 0 . So v u uy v u ux = u y 0 on D1 . By the C-R equations vx = v y 0 on D1 . So f is a constant function on each simply
≡
− |
≡ −
|
≡
−
≡
7
|
|
|
| { ∈
}
connected component of D 1 . Since D 2 := z D : f (z) = 0 has no accumulation points in D (see Theorem 2.13), f must be identical to the same constant throughout D. For (iv), we assume arg f (z) θ. Then g(z) := e −iθ f (z) is also holomorphic and f is a constant function if and only if g is a constant function. So without loss of generality, we can assume arg f (z) 0. Then Imf (z) 0. By C-R equations, Ref (z) is a constant too. Combined, we can conclude f is identically equal to a constant on D.
{ ∈
}
≡
≡
≡
22. Proof. The tangent vector a(x, y) of the curve u(x, y) = c1 at point (x, y) is perpendicular to (ux , uy ); the tangent vector b(x, y) of the curve v(x, y) = c2 at point (x, y) is perpendicular to (vx , vy ). Since b(x, y), which (vy , vx ) (vx , vy ) and (ux , uy ) = (vy , vx ) by C-R equations, we must have a(x, y) means the curves u(x, y) = c 1 and v(x, y) = c2 are orthogonal.
− ⊥
−
⊥
23. (1) Proof. Since
x = r cos θ y = r sin θ
, we have ∂ ∂r ∂ ∂θ
It’s easy to see A −1 (θ, r) =
1 r
get
∂ ∂x ∂ ∂y
− − cos θ r sin θ
=
r cos θ r sin θ
sin θ r cos θ
:= A(θ, r)
∂ ∂x ∂ ∂y
.
sin θ . Writing the Cauchy-Riemann equations in matrix form, we cos θ ∂ ∂x ∂ ∂y
u =
0 1
−
1 0
∂ ∂x ∂ ∂y
v.
Therefore, under the polar coordinate, the Cauchy-Riemann equations become ∂ ∂r ∂ ∂θ
u = A(θ, r)
∂ ∂x ∂ ∂y
u = A(θ, r)
∂ ∂x ∂ ∂y
0 1 1 0
−
v = A(θ, r)
0 1 −1 A (θ, r) 1 0
−
∂ ∂r ∂ ∂θ
1 r
∂ ∂r ∂ ∂θ
v =
0 r
−
0
v.
Fix z = re iθ , then u(r + ∆r, θ) + iv(r + ∆r, θ) − [u(r, θ) + iv(r, θ)] ∂u −iθ ∂v r ∂u ∂v f (z) = lim = e + i e−iθ = +i . iθ ∆r →0 ∆r · e ∂r ∂v z ∂r ∂r
(2) Proof. If f (z) = R(cos φ + i sin φ), then u = R cos φ and v = R sin φ. So ur =
∂R ∂φ ∂R ∂φ ∂R ∂φ ∂R ∂φ cos φ R sin φ , vr = sin φ+R cos φ , uθ = cos φ R sin φ , vθ = sin φ+R cos φ . ∂r ∂r ∂r ∂r ∂θ ∂θ ∂θ ∂θ
−
−
Writing the above equalities in matrix form, we have ∂R ∂r ∂φ ∂r
∂R ∂θ ∂φ ∂θ
ur = A vr
uθ , = A vθ
cos φ , where A = sin φ
R cos φ
Therefore the C-R equations become A
∂R ∂r ∂φ ∂r
u = r = vr
0
− 1r
1 r
0
uθ = vθ
8
0
− 1r
1 r
0
∂R ∂θ ∂φ ∂θ
A
−R sin φ
.
.
Note A −1 =
1 R
1 r
R r
− −
R cos φ R sin φ . So A −1 sin φ R cos φ ∂R ∂r ∂φ ∂r
0
1 r
0
∂R ∂θ ∂φ ∂θ
R r
0
=
− Rr1
0
0
A =
1 Rr
, i.e.
0
∂R ∂r ∂R ∂θ
= =
. Therefore
R ∂φ r ∂θ
.
−Rr ∂φ ∂r
24.
u uy Proof. By the C-R equations, J = x = u x vy vx uy = u x ux vx ( vx ) = u 2x + vx2 = f (z) 2 . J is vx vy the area of f (D), see, for example, Munkres [4], 21 The volume of a parallelepiped. 25. (1)
−
· − · −
|
|
§
Proof. We can parameterize γ with γ (t) = t + ti (0
≤ t ≤ 1). So
1
xdz =
t(dt + idt) =
0
γ
1 1 + i. 2 2
(2) Proof.
| − || z
γ
1 dz
2π
|
=
0
√ 2
=
2π
it
|e − 1|dt = π 4
− 34 π
0
√ 2 − 2cos tdt = √ 2
√ |sin θ − cos θ| dθ = 2
π 4
(cos θ
− 34 π
2π
− − − sin
0
π 4
t 2
cos
π 4
t 2
− sin θ) dθ4.
(3) Proof.
γ
dz z
−a
2π
=
0
iReit dt = 2πi. Reit
26. Proof. We first recall sinh z =
ez e−z 2
−
and coshz =
cos(x + iy) = cosh y cos x
ez +e−z . 2
Then straightforward calculations show
− i sinh y sin x, sin(x + iy) = cosh y sin x + i sinh y cos x.
By this result, we have sin iz =
ei(iz)
− e−i(iz) = e−z − ez = sinh z, cos iz = ei(iz) + e−i(iz) = e−z + ez = cosh z, 2i
2i
2
eiz + e−iz eiz − e−iz (sin z) = = = cos z.
2i
9
2
2
dt
and cos z1 cos z2 sin z1 sin z2 eiz1 + e−iz1 eiz2 + e−iz2 eiz1 e−iz1 eiz2 e−iz2 2 2 2i 2i − − i(z1 +z2 ) i(z1 +z2 ) i(z1 −z2 ) i(z1 −z2 ) e +e +e +e e i(z1 +z2 ) + e−i(z1 +z2 ) ei(z1 −z2 ) 4 4 − i(z1 +z2 ) i(z1 +z2 ) 2e + 2e 4 cos(z1 + z2 ).
−
=
−
−
=
−
− −
−
= =
− e−i(z −z ) 1
2
27. Proof. sin i = sinh1. cos(2 + i) = cosh 1 cos 2 i sinh1sin2. i sinh1 cosh1 cos 2 tan(1 + i) = sin1cos1+ . cosh2 1cos2 1+sinh2 1sin2 1 − i kπ 2 2 = e [cos(log 2) + i sin(log 2)], k Z. π +2 − i i = e 2 kπ , k Z. ( 1)2i = e −2(2k+1)π , k Z. log(2 3i) = log 13 + i[2kπ arctan 32 ], k
−
−
∈
∈ √ ∈
−
√ i( π arccos 14 (3 + i) = 3+i± 10e 42
−
1 arctan 3 ) 2 4
−
∈ Z.
.
28. π
Proof. e 2 i = i, e −
2π 3 i
1 2
=
−
√ 3 2
i. log i = (2kπ + π2 )i, k
∈ Z.
29. 2
x
Proof. z z = e z log z = e (x+iy) log(x+iy) = e 2 log(x
+y2 ) y(2kπ +arctan yx )
−
·ei[
y y 2 2 2 log(x +y )+x(2kπ +arctan x )]
, k
∈ Z.
30. 1
Proof. The root of z n = a are a n ei
| |
2kπ+arg a n
, k = 0, 1,
··· , n − 1.
31. (1) Proof. If u n (z) = Ref n (z) and v n (z) = Imf n (z), then by max un (z) um (z) , vn (z) vm (z) f n (z) f m (z) un (z) um (z) + vn (z) vm (z) , the Cauchy criterion for convergence for complex field is reduced to that for real numbers.
| ≤|
−
| |
−
{|
|
−
||
−
|} ≤ |
(2) Proof. Note
|
32. (i)
m n=k f n (z)
| ≤ M
1
m n=k an .
Proof. Since limn→∞ n n2 = e limn
→∞
ln n n2
So
∞ f (z) converges uniformly if n=1 n
= 1, Hadamard’s formula implies R = 1.
(ii) Proof. Since limn→∞ n
ln n n
= e limn
→∞
[ ln n ln n]
·
1, Hadamard’s formula implies R = 1.
(iii)
10
∞ a converges. n=1 n
−
∼ √ 2πn
Proof. By Stirling’s formula n! n! lim n→∞ nn
1 n
1
= lim e n ln n
n! nn
→∞
n n , e
we have
= lim e n [ln(n!)−ln 1
√ 2πn
( ne )n ]
n
→∞
e · nlim →∞
1 n ln
√ 2πn
( ne )n −ln n = e −1 .
So Hadamard’s formula implies R = e. (iv) Proof. Since limn→∞ n =
∞, Hadamard’s formula implies R = 0.
33. Proof. Let u(z) = Ref (z) and v(z) = Imf (z). Then f is uniformly convergent on a set A if and only if u and v are uniformly convergent on A. Since dz = dx + idy, we can reduce the theorem to that of real-valued functions of real variables. 34. (1) Proof. [f (z)e−z ] = f (z)e−z C = 1.
− f (z)e−z
= 0. So f (z) = Cez for some constant C . f (0) = 1 dictates
(2) Proof. f (z) = limω→0 f (z) = e z .
f (z +ω ) f (z ) ω
−
= limω→0 f (z) f (ωω)−1 = f (z)f (0) = f (z). By part (1), we conclude
35. (1) Proof. [ef (z) ] = f (z)ef (z) = 1. So e f (z) = 1. (2) Proof. It’s clear f (1) = 0. So f (z + zω) ω →0 zω
f (z) = lim Therefore
− f (z) = lim f (z(1 + ω)) − f (z) = 1 lim f (1 + ω) = 1 f (1) = 1 . ω
1 f (z) e z
zω
→0
z ω→0
ω
z
z
− 1z2 ef (z) + z1 ef (z)f (z) = 0.
=
This shows ef (z) = z. 36.
− ϕ(z2)] = (z1 − z2) 1 − z 1z . So ϕ(z) is univalent in any region that makes z1z2 = 1 whenever z 1 = z 2 . Since z 1 z2 = 1 implies |z1 | = |z1 | and arg z1 = − arg z2 , we conclude ϕ is univalent in any Proof. 2[ϕ(z1 )
1 2
of the four regions in the problem.
2
43. Proof. This is straightforward from Problem 24. 44.
11
Proof. We first use Abel’s Lemma on partial sum: for n > m, n
n
n
ak bk =
k=m+1
(S k
k =m+1
− S k−1)bk = S bbn−1 +
S k (bk
− bk+1) − S mbm+1.
k =m+1
Denote by K a bound of the sequence (S n )∞ K bn+1 , S m bm+1 K bm+1 , and n=1 , then S n bn+1 ∞ n n b k+1 ) K k=m+1 bk b k+1 . Since limn→∞ bn = 0 and b n+1 < , k=m+1 S k (bk n=1 bn n we conclude k=m+1 ak bk can be arbitrarily small if m is sufficiently large. So under conditions (1)-(3), ∞ a b is convergent. n=1 k k In the Dirichlet criterion, we require (1) (S n )∞ n=1 is bounded; (2) limn→∞ bn = 0; (3’) (bn )∞ n=1 is monotone. Clearly, (2)+(3’) imply (3), so the Dirichlet criterion is a special case of the result in current problem. In the Abel criterion, we require (1”) (S n )∞ n=1 is convergent; ∞ (2”) (bn )n=1 is bounded; (3”) (bn )∞ n=1 is monotone. ∞ Define bn = bn b, where b = limn→∞ bn is a finite number. Then (S n )∞ n=1 satisfies (1) and (bn )n=1 ∞ N satisfies (2) and (3’). By the Dirichlet criterion, n=1 an bn converges. Note N n=1 an bn = n=1 an bn +bS N . ∞ By condition (1”), we conclude n=1 an bn converges. Remark 1. The result in this problem is the so-called Abel-Dedekind-Dirichlet Theorem.
|
−
|≤
| −
|
|
−
|≤ |
| |
|≤ | | | − | ∞
45. (1) Proof. For any ε (0, 1 q ), there exists N ∞ a and ∞ (q + ε)n , we conclude n=1 n n=1
∈
| |
(2)
∈ ∞N, such that for any n ≥ N , |an|
−
1 n
< q + ε < 1. By comparing
n=1 an is absolutely convergent.
Proof. By definition of upper limit and the fact q > 1, we can find ε > 0 and infinitely many n, such that 1 ∞ an n > q ε > 1. Therefore limn→∞ an = and n=1 an is divergent.
| |
−
| | ∞
46.
| < q + ε < 1. So Proof. If q < 1, for any ε (0, 1 q ), we can find N N, such that for any n N , |a|an+1 n| ∞ ∞ aN +k aN (q + ε)k , k 0. Since k=1 (q + ε)k is convergent, we conclude n=1 an is also convergent. If q > 1, we have two cases to consider. In the first case, the upper limit is assumed to be a limit. Then we can find ε > 0 such that q ε > 1 and for n sufficiently large, an+1 > an (q ε) > an . This implies limn→∞ an = 0, so the series is divergent. In the second case, the upper limit is assumed not to be a limit. Then we can manufacture counter examples where the series is convergent. Indeed, consider ( k 0)
|
| ≤| |
∈ ∀ ≥
−
−
∈
|
Then 47.
| | | −
| |
≥
an =
≥
∞ a is convergent and lim n→∞ n=1 n
1 (k+1)2 , 2 (k+1)2 ,
an+1 an
= 2.
12
if n = 2k + 1 if n = 2k + 2.
Proof. We choose ε > 0, such that limn→∞ n
|
an+1 an
∈ N, such that for any n ≥ N 1, n | aa
exists N 1 Then
bn+1 nε = = 1 bn (n + 1)1+ε
n+1 n
−
So there exists N 2 n
≥ N ,
an+1 an
<
bn+1 . bn
n+1
1+ε
bn+1 bn
=1
−
1 (n + 1)2
.
Define N = max N 1 , N 2 , then for any
{
· ··
}
aN +k bN +k , < . aN +k−1 bN +k−1
|aN +k | ≤ |abN N | bN +k .
∞ b is convergent, we conclude n=1 n
48. (1)
1+ε + O n+1
ε − 1+2 n+1 .
>1
1+ε
n
aN +1 bN +1 aN +2 bN +2 < , < , aN bN aN +1 bN +1
Multiply these inequalities, we get
In particular, we have
Since
1 n+1
∈ N, such that for n ≥ N 2,
| − 1 < −1 − 2ε. By the definition of upper limit, there | − 1 < −1 − 2ε, i.e. |a|a | | < 1 − 1+2n ε . Define b n = n 1 .
∞ a must also converge. n=1 n
1
Proof. Since (an )∞ n=1 monotonically decreases to 0, for n sufficiently large, ln an < 0 and limn→∞ 1 n
ln an n
So limn→∞ an = e limn
→∞
ln an n
≤ e0 = 1. By Hadamard’s formula, R ≥ 1.
≤ 0.
(2) Proof. The claims seems problematic. For a counter example, assume R > 1 and a n = θ = 0, we have
N
N
N
N
1 Rn .
If z = Re iθ with
θ θ sin ( N +1) cos N2θ sin ( N +1) sin N2θ 1 n inθ 2 2 an z = R e = cos nθ + i sin nθ = +i . θ θ n R sin sin 2 2 n=0 n=0 n=0 n=0
So 49.
n
·
∞ a zn is not convergent. n=0 n
Proof. By definition of uniform convergence, for any ε > 0, there exists N ∞ a z z n < ε, z D. In particular, for any z∗ ∂D, by letting z 0 n=m n z0 n ε. That is, for this given ε, we can pick up N , such that for any m ¯ This is exactly the definition of uniform convergence on D. ¯ z D.
| ≤ ∀ ∈
| || − |
∀ ∈
∈ N, such that for any m ≥ N , ∗ → z , we∞ get ∞n=m |ann||z∗ − ≥ N , n=m |an||z − z0| ≤ ε,
∈
50. Proof. We choose a sequence (rn )∞ (0, 1) and limn→∞ rn = 1. Define β n = n=1 such that rn f (rn z0 ). We show limn→∞ β n = 0. Indeed, we note
∈
| | n
|β n|
=
≤ 1
ak z0k
k =0 n
≤
∞
−
k =0
n
ak (rn z0 )k =
k=0
ak (1
− rnk )z0k
| − rn)(1 + rn + ··· + rnk−1) +
ak (1
k=0 n
| − rn)k + ε nn+1 1 −1 rn ,
ak (1
k=0
an+1 = 1 is redundant. an
It seems the condition limn→∞
13
∞
−
∞ k
k=n+1
k =n+1
ak |rnk | n
ak (rn z0 )k
n k k=0 ak z0
−
where εn = k≥n k ak . We choose rn in such a way that lim n→∞ n(1 εn+1 limn→∞ n(1 −rn ) = 0 since lim n→∞ nan = 0. By Problem 18 (2), we have
| |
n
|
ak (1
| − rn)k = (1 − rn)n
k=0
n k =0
·
− rn) = 1, e.g.
rn = 1
− n1 .
Then
|ak |k → 0, as n → ∞.
n
Combined, we conclude limn→∞ β n = 0. Remark 2. The result in this problem is a special case of the so-called Landau’s Theorem, see, for example, Boos [1], 5.1 Boundary behavior of power series.
§
2
Cauchy Integral Theorem and Cauchy Integral Formula
1. Proof. (i) The integral is equal to 2 πi sin i = π(e−1 e). (ii) The integral is equal to 2eπi. (iii) For sufficiently small ε > 0, the integral is equal to
|z−π|=ε
cos z (z + π)(z
·
−
cos z (z + π)(z
− π) dz +
(iv) The integral is equal to 2πi
1
z 3
−
|z+π|=ε (n−1)
z =1
−
cos π cos( π) dz = 2πi + = 0. π) 2π 2π
1)! = 2πi − (n2− = n
−
−
−(2n−1)! πi.
·
n−1
(v) For sufficiently small ε > 0he integral is equal to
1 dz 1 dz + = 2πi 2 2 i)(z + 4) (z + i) |z−i|=ε (z + i)(z + 4) (z i)
1 1 + = 0. ( 2i) 3 2i 3
· − − · · (vi) By the Fundamental Theorem of Algebra, the equation z 5 − 1 = 0 has five solutions z 1 , z2 , z3 , z4 , z5 . For any i ∈ {1, 2, 3, 4, 5}, we have z5 − 1 z 5 − zi5 5 (zi − z1 ) ··· (zi − zi−1 )(zi − zi+1 ) ··· (zi − z5 ) = lim = lim = 5zi4 = . z →z z − zi z →z z − zi zi |z+i|=ε (z −
·
·
i
i
So for sufficiently small ε > 0, the integral is equal to 5
i=1
|z−zi |=ε (z − z1 )(z −
5
= 2πi
i=1 6
= 2πi
i=1
(zi
− z1) ··· (zi −
dz z2 )(z z3 )(z
−
− z4)(z − z5)
1 zi−1 )(zi
− zi+1) ··· (zi − z5)
zi 5
= 0,
where the last equality comes from the fact that the expansion of (z z1 ) (z z5 ) is z 5 1 and (z1 + +z5 ) is the coefficient of z 4 . (vii) If a , b > R, the integral is equal to 0. If a > R > b , the integral is equal to (b2−πi a)n . If
− ··· − − | | || (n−1) = 2πi · ( −1)n−1 (a − b)−n .
| | | | |b| > R > |a|, the integral is equal to 2πi z−1 b integral is equal to 2πi (b−1a) + (−1)n−1 (a−1b) = 0.
n
n
14
z =a
−
···
If R > a , b , the
| | | |
(viii) Denote by z 1 , z2 , z3 the three roots of the equation z 3 2πi = 2πi
1
− 1 = 0. Then the integral is equal to
1
1
− z2)(z1 − z3)(z1 − 2)2 + (z2 − z1)(z2 − z3)(z2 − 2)2 + (z3 − z1)(z3 − z2)(z3 − 2)2 + z1 z2 z3 12 + + . − 3(z1 − 2)2 3(z2 − 2)2 3(z3 − 2)2 49 (z1
− z3
2. Proof. 1 2πi
ezζ dζ 1 = n ζ 2πi C ζ
·
∞ zk 1 dζ ∞ zk 1 · = k! · 2πi k! ζ n−k ζ
C k =0
k =0
1
ζ n−k
C
·
dζ z n = , ζ n!
where the last equality is due to Cauchy’s integral formula. We need to justify the exchange of integration and infinite series summation. Indeed, for m > n, we have
· − · · · || || ≤ || || ≤ || | | || | | · − · ≤ · || · · · | | m
k =0
1 2πi
=
1 2π 1 2π
zk k!
1 2πi
1
C
∞
C k=m+1
∞
C k=m+1
∞
C k=m+1
ζ n−k
dζ ζ
1 2πi
∞ zk 1 dζ k! ζ n−k ζ
C k=0
zk 1 dζ k! ζ n−k ζ
zζ k dζ k! ζ n+1
M k dζ , k! ζ n+1 k
∞
M where M = z maxζ ∈C ζ . Since k=m+1 M k! is the remainder of e , for m sufficiently large, can be smaller than any given positive number ε. So for m sufficiently large, m
k =0
This shows 3.
zk k =0 k !
∞
zk k!
1 2πi
1 2πi
1 C ζ n−k
1
C
ζ n−k
dζ ζ
1 2πi
dζ ζ converges
to
∞ zk 1 dζ k! ζ n−k ζ
ε
zk 1 k=0 k! ζ n−k
dζ ζ .
C k=0
1 2πi
C
∞
1 2π
C
M k k=m+1 k !
∞
dζ . ζ n+1
g (ζ ) f (ζ ) 1 − = ζ − z1 . If z < 1, we apply Cauchy’s integral formula to 2πi |ζ |=1 ζ −z dζ and apply g (ζ ) (ζ ) 1 Cauchy’s integral theorem to 21πi |ζ |=1 ζ g− an analytic function of 1 dζ (regarding z as a parameter and ζ − z1 z (ζ ) f (ζ ) ζ ) . If |z | > 1, we apply Cauchy’s integral theorem to 21πi |ζ |=1 f ζ −z dζ (regarding z as a parameter and ζ −z (ζ ) an analytic function of ζ ) and apply Cauchy’s integral formula to 21πi |ζ |=1 ζ g− 1 dζ . z
Proof. We note
4.
g (ζ ) zζ 1
Proof. For z z : z < 3 , by Cauchy’s integral formula, f (z) = 2πi(3z 3 + 7z + 1). So f (1 + i) = 2πi[6(1 + i) + 7] = 12π + 26πi.
∈ { | | −
}
5.
15
(1)
1
1
z =2
Proof. We note
2π
1 dz (z + )2n = z z |z|=1
(e + e−iθ )2n idθ = 22n i iθ
0
2π
cos2n θdθ.
0
On the other hand, 1 1 (z + )2n = z z
·
Therefore 2π
2n
cos
0
2n
2n
C 2kn z k (z 1 )2n k
− · z −1 =
−
k=0
k=0
C 2kn z −2n+2k−1
.
1 dz 1 C 2nn dz 2πC 2nn 1 3 5 θdθ = 2n (z + )2n = 2n = = 2π n 2 2 i |z|=1 z z 2 i |z|=1 z 2 2 4 6 1
− 1) . · · ·· ·· · ······(2n · · 2n
6. Proof. This is straightforward from Cauchy’s integral formula. 7. Proof. For any R > 0, we have f (n) (0) =
(n)
f
(0) 1 = n! 2π
2π
0
n! 2πi
f (ξ ) |z|=R ξn+1 dξ . So
|f (Reiθ )| dθ ≤ 1 Rn 2π
Let R = n, we get the desired inequality. R
e Remark 3. Note h(R) = ln R n = R
2π
iθ Me|Re |
Rn
0
− n ln R takes minimal value at n.
e R dθ = M n . R
This is the reason why we choose
R = n, to get the best possible estimate. Another perspective is to assume f (z) = e z . Then
√ 2πn1 n
(e)
n
= √ 21πn
e n n
R
f (n) (0) n!
e by Stirling’s formula. So the bound M R n is not “far” from the best one.
=
1 n!
∼
8. (1) Proof. For any z D2 , we can find R > 0 sufficiently large, so that z is in the region enclosed by γ and ζ : ζ z = R . By Cauchy’s integral formula
{ | − |
}
∈
1 f (ζ ) 1 f (ζ ) f (z) = dζ + dζ. 2πi −γ ζ z 2πi |ζ |=R ζ z
−
−
So 1 2πi
f (ζ ) dζ = z γ ζ
−
= We note
1 f (ζ ) f (z) + dζ 2πi |ζ |=R ζ z 1 f (ζ ) f ( ) 1 f ( ) f (z) + dζ + dζ. 2πi |ζ |=R ζ z 2πi |ζ |=R ζ z
− −
≤
1 f (ζ ) f ( ) dζ 2πi |ζ |=R ζ z
− ∞ −
and Cauchy’s integral formula implies
−
sup ζ D(z,R )
∈
− ∞ −
∞ −
∞·
16
∞ −
|f (ζ ) − A| → 0, as R → ∞,
1 f ( ) dζ = f ( ) 1 = A. 2πi |ζ |=R ζ z
So by letting R
→ ∞ in the above equality, we have 1 2πi
f (ζ ) dζ = z γ ζ
−f (z) + A.
−
(ζ ) If z D 1 , h(ζ ) = f ζ − z is a holomorphic function in D 2 , with z a parameter. So for R sufficiently large, Cauchy’s integral theorem implies
∈
1 f (ζ ) 1 f (ζ ) 0= dζ + dζ. 2πi −γ ζ z 2πi |ζ |=R ζ z
(2) Proof. We note
z zζ ζ 2
− =
−
−
f (ζ ) dζ = A. γ ζ z
1 2πi
An argument similar to (1) can show
−
1 ζ
− ζ −1 z . For R sufficiently large, Cauchy’s integral theorem implies 1 f (ζ ) 1 f (ζ ) 0= dζ + dζ. 2πi −γ ζ 2πi |ζ |=R ζ
So
1 2πi
f (ζ ) dζ = 21πi γ ζ
|ζ |=R
f (ζ ) ζ dζ .
1 2πi
Similarly,
f (ζ ) dζ = z γ ζ
−
Therefore 1 2πi
zf (ζ ) 1 dζ = 2 ζ 2πi γ zζ
−
f (ζ ) |ζ |=R ζ −z dζ − f (z) z ∈ D 2 f (ζ ) z ∈ D 1 . |ζ |=R ζ −z dζ
1 2πi 1 2πi
γ
f (ζ ) ζ
f (ζ ) dζ = ζ z
− −
f (z) z 0 z
∈ D2 ∈ D1.
9. Proof. Since f (z) = 0 in D, 1/f (z) is holomorphic in D. Applying Maximum Modulus Principle to 1/f (z), we conclude f (z) achieves its minimum on ∂D. Applying Maximum Modulus Principle to f (z), we conclude ¯ By Chapter f (z) achieves its maximum on ∂ D. Since f (z) M on ∂ D, f (z) must be a constant on D. 1, exercise problem 21 (iii), we conclude f (z) is a constant function.
|
|
|
|
|
|≡
|
|
10. Proof. For R large enough, a , b < R. So we can find ε > 0 such that (assume a = b)
| | | |
|z|=R (z −
f (z) a)(z
− b)
dz =
|z−a|=ε (z −
f (z) a)(z
− b)
dz +
|z−b|=ε (z −
f (z) a)(z
− b)
dz =
2πi [f (a) a b
−
Meanwhile, denote by M a bound of f , we have
|z|=R (z −
f (z) a)(z
− b)
dz
≤
M
|dz|
|z|=R |(z − a)(z − b)|
|dz| |z|=R (R − |a|)(R − |b|) 2πM R = → 0, as R → ∞. (R − |a|)(R − |b|) πi Combined, we conclude 0 = a2− b [f (a) − f (b)], ∀a, b ∈ C. This shows f is a constant function. ≤
M
17
− f (b)].
11. Proof. By Cauchy’s integral formula,
|f (n)(0)| ≤
n! 2π
|z|=r
≤
f (ξ ) dξ ξ n+1
n! 2π
2π
0
1 1 r rn
− dθ =
n! . rn (1 r)
−
12. Proof. We note Φ(z) lim z →z0 z
− Φ(z0) = 1 lim − z0 2πi z→z
0
γ
(ζ
−
ϕ(ζ ) 1 dζ = z)(ζ z0 ) 2πi
−
ϕ(ζ ) dζ, z 0 )2 γ (ζ
−
where the exchange of limit and integration can be seen as an application of Lebesgue’s Dominated Convergence Theorem (We can find ρ > 0 so that z in a neighborhood of z0 , dist(γ, z) ρ. Then (ζ −z)(1ζ −z0 ) ) ρ12 ). This shows Φ is holomorphic in any region D which does not contain any point of γ . The formula for n-th derivative of Φ can be proven similarly and by induction. For details, we refer the reader to Fang [3], Chapter 4, 3, Lemma 2. Remark 4. The result still holds if ϕ is piecewise continuous on γ . See, for example, Fang [3], Chapter 7, proof of Theorem 7.
∀
≥
≤
§
13. Proof. Since f (z) = 0 in D, 1/f (z) is holomorphic in D. Applying Maximum Modulus Principle to 1/f (z), we conclude f (z) does not achieve its minimum in D unless it’s a constant function. Applying Maximum Modulus Principle to f (z), we conclude f (z) does not achieve its maximum in D unless it’s a constant function. Combined, we have m < f (z) < M for any point z D.
| |
|
|
∈
14. Proof. Define f (z) = pnz(nz) and g (z) = f ( 1z ). Then g(z) is holomorphic and bounded on D(0, 1) 0 , since limz→0 g(z) = limz→∞ pnz(nz) is a finite number and g(z) is clearly continuous in a neighborhood of ∂ D(0, 1). By Theorem 2.11, Riemann’s Theorem, g(z) can be analytically continued to D(0, 1). By a corollary of Theorem 2.18, Maximum Modulus Principle, the maximum value of g(z) can only be reached on ∂ D(0, 1). So g(z) max |z|=1 g(z) = max|z|=1 f (z) = max|z|=1 pn (z) M , z D(0, 1), i.e. f (z) = | p|nz(nz|)| M for 1 z < .
\{ }
|
|≤ ≤ | | ∞
|
|
|
|
|
| | | ≤ ∀ ∈
|
|
≤
15. Rz ) Proof. Define g(z) = f (M (z D(0, 1)) and apply Theorem 2.19, Schwarz Lemma, to g(z), we have z| z g(z) z ( z D(0, 1)) and g (0) 1. So for any z D(0, R), f (z) = f (R R ) = M g( zR ) M |R and M M f (0) = g (0) R . R
| |
| ≤ | | ∀ ∈ | | | · ≤
|
∈
|≤
∈
·
16. Proof. Define ρ = dist(C V, K ). Then ρ
\
(n)
|f Define c n =
n! 2π
(z)
dξ and ∂V ρn+1
| |
|≤
n! 2π
∈ (0, ∞). By Theorem 2.7, for any z ∈ K , we have |f (ξ )| |dξ | ≤ n! |dξ | · sup |f (z)|. n+1 2π ∂V ρn+1 z∈V ∂V |ξ − z |
take supremum on the left side of the above inequality, we have sup f (n) (z) z K
∈
|
| ≤ cn sup |f (z)|. z ∈V 18
≤
17.
|
| ∞ |
−
||
|≤
| |
Proof. We have to assume γ has finite length, that is, L(γ ) := γ dz < . Then by the definition of uniform ε convergence, for any ε > 0, there exists n0 such that n n0 , supz∈D f (z) f n (z) L(γ ) . So for any n n 0 , ε f (z)dz f n (z)dz f (z) f n (z) dz dz = ε. γ γ γ γ L(γ )
∀ ≥
≥
Therefore limn→∞ 18.
≤ |
−
f (z)dz = γ n
f (z)dz.
γ
−
|≤
α Proof. We note the linear fractional transformation ω(z) = zz− +α maps z : Rez > 0 to D(0, 1). So h(z) = ω(f (z)) maps D(0, 1) to itself (see, for example, Fang [3], Chapter 3, 7) and h(0) = 0. By Schwarz Lemma, h(z) = z and h (0) 1. This is equivalent to
{
| | |
|
|≤
f (z) α f (z) + α
}
§
− ≤ |z|, and |f (0)| ≤ 2α.
19. Proof. The linear fractional transformation ω(z) = ω(f (z)) =
{ z
: Rez < A to D(0, 1). So h(z) =
} to itself and h(0) = 0. By Schwarz Lemma, |h(z)| ≤ |z |. So ∀z ∈ D(0, 1) |. |f (z)| ≤ |z| · |f (z) − 2A| ≤ |z||f (z)| + 2A|z|, i.e. |f (z)| ≤ 2A1 −|f (z) |z|
f (z ) f (z ) 2A maps D(0, 1)
−
z z 2A maps
−
20. (1) Proof. We note ez + e−z + 2 cos z 4
=
1 z 1 [e + e−z + eiz + e−iz ] = 4 4
The radius of convergence is
∞ 1
k =0
k!
k
k
k
k
[z + ( z) + (iz) + ( iz) ] =
−
−
∞ z4m
(4m)! m=0
.
∞.
(2) Proof. By induction, it is easy to show is
1 (1 ζ )3
−
=
∞
(n+1)(n+2) n ζ , n=0 2
1 (1 ζ )3
−
(n)
=
(n+2)! 2! (1
− ζ )−(n+3). So the Taylor expansion of (1−1ζ )
3
with radius of convergence equal to 1. This implies
∞
ζ + 4ζ 2 + ζ 3 = ζ + 7ζ 4 + (3n2 (1 ζ )3 n=3
−
− 3n + 1)ζ n. ∞
Replace ζ with z 2 , we can get the Taylor expansion at z = 0: z 2 + 7z 8 + n=3 (3n2 3n + 1)z 2n , and the 1 radius of convergence is equal to 1 (this can be verified by calculating limn→∞ (3n2 3n + 1) n = 1).
(3)
19
− −
(n)
Proof. By induction, it is easy to show (1 ζ )−4 = (n+3)! ζ )−(n+4) . So for ζ D(0, 1), (1 3! (1 ∞ (n+1)(n+2)(n+3) ζ n . Replace ζ with z 5 , we get for z D(0, 1) n=0 6
−
−
∈
∈
− ζ )−4 =
∞
1 (n + 1)(n + 2)(n + 3) 5n = z . (1 z 5 )4 n=0 6
−
Therefore (1 − z −5 )−4 =
∞
z 20 (n + 1)(n + 2)(n + 3) 5n+20 = z . 5 4 (1 z ) 6 n=0
−
By Hadamard’s formula, the radius of convergence is 1. (4) Proof. By induction, we can show (1 + z)−2
(n)
= ( 1)n (n + 1)!(1 + z)−(n+2) . So z
∀ ∈ D(0, 1),
−
∞
1 = ( 1)n (n + 1)z n . (1 + z)2 n=0
−
It is also easy to verify that 1 (z + 1) 2 (z
− 1) = −
1 1 + 4(z + 1) 4(z 1)
− −
1 = 2(z + 1) 2
∞
1 ( z)n 4 n=0
− −
−
∞
1 zn 4 n=0
−
∞
1 ( 1)n (n + 1)z n . 2 n=0
−
Therefore z6 (z + 1)(z 2
− 1) = −
z6 4
∞
−
n
n
[( 1) + 1 + 2 ( 1) (n + 1)] z =
·−
n=0
∞
1 [( 1)n (2n + 3) + 1] z n+6 . 4 n=0
− −
n
By Hadamard’s formula, the radius of convergence is 1. 21. (1) Proof. We note π 1 f (re iθ ) 2 dθ 2π −π
|
|
= = =
∞
| | π 1 2π −π
1 2π
∞
m,n=0
∞
n=0 π
an rn einθ
∞
·
a ¯m rm e−imθ
m=0
dθ
an a ¯m rn+m ei(n−m)θ dθ
−π
an a ¯m rn+m δ nm
m,n=0
=
∞
an 2 r 2n ,
n=0
1 if m = n and the exchange of integration and summation is justified by Theorem 1.14 0 if m = n (Abel Theorem) and Problem 17. where δ mn =
(2)
20
Proof. In the result of part (1), multiply both sides by r and integrate with respect to r from 0 to 1, we have
∞
1
| | | an
2
r
0
n=0
∞ |an |2 = n=0 n+1
So 22.
2n+1
1 π
∞
1
an 2 1 = dr = 2n + 2 2π n=0
|
|
π
1 f (re ) rdrdθ = 2π −π
| 0
iθ
2
|
|
f (z) 2 dA.
|
D
f (z) 2 dA.
D
|
¯ R), then z ¯ 1), v(z) := u(zR) is Proof. Suppose u is harmonic on D(0, R) and continuous D(0, D(0, ¯ 1). By formula (2.31), for any z D(0, R), harmonic on D(0, 1) and continuous on D(0,
∀ ∈ ∈
2 z 2 2π 2 − 1 iτ iτ R − |z | R v(e ) dτ = u(Re ) z¯ iτ 2 2π 0 |R − ¯zeiτ |2 dτ 0 1− R e 2π 2π R2 − |z |2 1 R2 − |z |2 iτ u(Re ) |R − ze−iτ |2 dτ = 2π 0 u(ζ ) |ζ − z|2 dτ. 0
z 1 u(z) = v( ) = R 2π 1 2π
=
2π
1
This is formula (2.32). To verify formula (2.33), note
− ¯z) = Re R2 − |z|2 + R(ze−iτ − ¯zeiτ ) = R2 − |z|2 . − | |Reiτ − z|2 |Reiτ − z|2
Reiτ + z (Reiτ + z)(Re−iτ Re iτ = Re Re z Reiτ z 2
−
|
Therefore, u(z) =
1 2π
= Re
2π
R2 z2 1 u(Re ) dτ = 2 iτ Re z 2π
0
iτ
−| | | − |
2π
1 ζ + z dζ u(ζ ) . 2πi |ζ |=R ζ z ζ
−
−
0
2π
Reiτ + z 1 u(Re )Re iτ dτ = Re Re z 2π iτ
0
Reiτ + z u(Re ) iτ dτ Re z iτ
−
This is formula (2.33). 23. Proof. On ∂ D(0, 1), ( 6z) (z 4 6z + 3) = z 4 + 3 4 < 6z . By Rouch´e Theorem, z 4 6z +3 and 6z have the same number of zeros in D(0, 1), which is one. On ∂D(0, 2), z 4 (z 4 6z +3) = 6z 3 15 < z 4 . So z 4 6z + 3 and z 4 have the same number of zeros in D(0, 2), which is four. Combined, we conclude z 4 6z + 3 = 0 has one root in D(0, 1) and three roots in the annulus z : 1 < z < 2 .
| − − −
| |
|≤
|− |
−
−
− | − − | | − | ≤ { | | }
− | |
24. Proof. On ∂ D(0, 1), ( 5z 4 ) (z 7 5z 4 z + 2) = z 7 z + 2 4 < 5z 4 . So z 7 have the same number of zeros in D(0, 1), which is four (counting multiplicity).
| −
− −
−
| | −
|≤
|− |
− 5z4 − z +2 and −5z4
25. R and z Proof. Let P (z) = z 4 + 2z 3 2z + 10. We note P (z) = (z 2 1)(z + 1)2 + 11. If z 1, 2 1, P (z) 1 (1 + 1) + 11 = 7. So P (z) = 0 has no root on the real R and z P (z) 11; if z axis. If z = iy with y R, we have P (iy) = y 4 + 10 2iy(y2 + 1) = 0. So P (z) = 0 has no root on the imaginary axis. Consider the region D enclosed by the curve γ = γ 1 γ 2 γ 3 , where (R is a positive number) γ 1 = z : 0 Rez R, Imz = 0 , γ 2 = z : z = R, arg z [0, π2 ] , and γ 3 = z : 0 Imz R, Rez = 0 .
≥
{
∈
≤
∈
≤
− | | ≤
}
−
≥ − ·
{ | |
On γ 1 , ∆γ 1 arg P (z) = 0. On γ 2 , P (z) = z 4 1 +
2z
− ∈
3
}
∪ ∪
{
≤
∈
| | ≥
≤
}
−24z+10 . So ∆γ arg P (z) = 4 · π + o(1) = 2π + o(1) 2 2 z
→ ∞). On γ 3, ∆γ arg P (z) = arg P (0) − arg P (iR) = arg10 − arg(R4 + 10 − 2iR(R2 + 1)) = 0 − +1) arg 1 − 2i RR(R+10 = o(1) (R → ∞). Combined, we have ∆γ arg P (z) = 3k=1 ∆γ arg P (z) = 2π + o(1) (R → ∞). (R
2
4
3
21
k
By The Argument Principle, P (z) = 0 has only one root in the first quadrant. The root conjugate to this root must lie in the fourth quadrant. The other two roots are conjugate to each other and are located in the second and third quadrant. So one must lie in the second quadrant and the other must lie in the third quadrant. Remark 5. The above solution can be found in Fang [3], Chapter 6, 3 Example 5.
§
26. Proof. On ∂ D(0, 1), ( 8z +10) (z 4 8z +10) = z 4 = 1 < 10 8 z 8z + 10 . So z 4 8z +10 and 8z + 10 have the same number of zeros in D(0, 1), which is zero. On ∂ D(0, 3), z 4 (z 4 8z + 10) = 8z 10 < 34 < z 4 . So z 4 8z + 10 and z 4 have the same number of zeros in D(0, 3), which is four (counting multiplicity). So z 4 8z + 10 = 0 has no root in D(0, 1) and four roots in the annulus z : 1 < z < 3 .
| − − − | | | | − | || ≤ |− | − − | | − | | | − − { | | }
−
|
−
27. Proof. If a > e, on ∂ D(0, 1), az n (az n ez ) = ez e |z| = e < az n . By Rouch´e Theorem, e z az n have the same number of zeros in D(0, 1), which is n.
| −
− | | | ≤
| |
− azn and
28. Proof. On ∂ D(0, 1), z (z f (z)) = f (z) < 1 = z . So z f (z) and z have the same number of zeros in D(0, 1), which is one. So there is a unique fixed point of f in D(0, 1). Remark 6. This is a special case of Brower’s Fixed Point Theorem.
| − −
| |
|
| |
−
29. (1) Proof. Re(ze z ) = e x (x cos y
− y sin y).
(2) Proof. Let u(x, y) = ax 3 + bx2 y + cxy 2 + dy 3 . Then ∆u = (6a + 2c)x + (2b + 6d)y. So u is harmonic if and only if 3a + c = 0 and b + 3d = 0. So the most general harmonic function of the form ax 3 + bx2 y + cxy 2 + dy 3 is ax 3 3dx2 y 3axy 2 + dy 3 .
−
−
30. 2
∂ ∂ ∂ Proof. Let u(r, θ) = ln(1 2r cos θ + r2 ). Then using the fact ∆ = 1r ∂r r ∂r + r12 ∂θ 2 , we can verify ∆u = 0. 2π 1 So by the mean-value property of harmonic functions, we have 2π 0 u(r, θ)dθ = u(0, 0) = 0. Since cos θ π 2π is an even function of θ, it’s not hard to show 0 ln(1 2r cos θ + r 2 )dθ = 2 0 ln(1 2r cos θ + r 2 )dθ. So π 2r cos θ + r2 )dθ = 0. 0 ln(1
−
−
−
−
31. Proof. Let M = supz∈C u(z) . Then z C and R > z , the holomorphic function defined in formula (2.34) satisfies 2π 1 Reiθ + z M 2π R + z f (z) u(Reiθ ) dθ dθ M, as R . 2π 0 Reiθ z 2π 0 R z
|
|
|≤
|
∀ ∈
|
|
| |
−
≤
| | → −| |
→ ∞
So f (z) is bounded on the entire complex plane, and by Liouville’s Theorem, f is a constant function. Therefore u = Re(f ) is a constant function. 32.
22
Proof. If the arc is ∂ D(0, 1), the desired harmonic function u 1. So without loss of generality, we assume the arc γ = z ∂D(0, 1) : θ 1 arg z θ 2 with 0 < θ2 θ1 < 2π. For n sufficiently large, we consider γ n = z ∂ D(0, 1) : θ 1 n1 arg z θ 2 + n1 . By the Partition of Unity Theorem for R1 , we can find ϕn C (∂D(0, 1)) such that 0 ϕ n 1, ϕn 1 on γ , and ϕn 0 on 2π ∂D(0, 1) γ n . Define u n (z) = 0 P (ζ, z)ϕn (ζ )dθ where P ( , ) is the Poisson kernel and ζ = Re iθ . Then u n is the unique solution of the Dirichlet problem with boundary value ϕ n . So ϕ n is harmonic in D(0, 1). It is easy to see that if (ϕn )n uniformly converges to the indicator function 1γ (z), (un )n uniformly 2π converges to u(z) := 0 P (ζ, z)1γ (ζ )dθ on D(0, ρ) (ρ (0, 1)). Since each u n satisfies the local mean value property, u must also satisfies the local mean value property in D(0, 1) and is continuous. By Theorem 2.22, u is harmonic on D(0, 1). This suggests the harmonic function we are looking for is exactly u. What is left to prove is that for any z0 ∂D(0, 1), lim|z|=1,z→z0 u(z) = 1γ (z0 ). Indeed, in general, we have the following classical result: Suppose function ϕ is piecewise continuous on ∂D(0, 1), then the 2π function u(z) := 0 P (ζ, z)ϕ(ζ )dθ is harmonic in D(0, 1) and for any continuity point z0 of ϕ, we have limz→z0 ,|z|<1 u(z) = ϕ(z0 ). For the proof of continuity on the boundary, we refer to Fang [3] Chapter 7, Theorem 7.
{ ∈
≤
≤ } { ∈ ∈
\
≡ − − ≤ ≤ } ≤ ≤ ≡ ··
≡
∈
∈
33. Proof.
∀ε > 0, there exists N , such that for any n ≥ N , p ∈ N, we have n+ p
≤ sup
ζ ∂U
∈
By Maximum Modulus Principle,
f k (ζ ) < ε.
k=n+1
n+ p
sup
¯ z U k=n+1
∈
f k (z)
n+ p
sup
ζ ∂U
∈
f k (ζ ) < ε.
k=n+1
∞ ¯ So Cauchy criterion implies n=1 f n (z) converges uniformly on U . Remark 7. This result is the so-called Weierstrass’ Second Theorem.
34. f (z0 ) f (z ) ¯ Proof. Define g(z) = f (zz)− −z0 = z−z0 , z D(0, R) z0 . Then g(z) can be analytically continued to ¯ R). By Cauchy’s integral D(0, R) (Theorem 2.11, Riemann Theorem) and is therefore continuous on D(0, formula, we have f (0) 1 g(ζ ) 1 f (ζ ) = g(0) = dζ = dζ. z0 2πi |ζ |=R ζ 2πi |ζ |=R ζ (ζ z0 )
∀ ∈
−
Therefore
\{ }
≤ f (0) z0
1 2π
2π
0
−
|f (Reiθ ) dθ ≤ M , |Reiθ − z0| R − |z0|
which is equivalent to R f (0) (M + f (0) ) z0 . Remark 8. The point of this problem is to give an estimate of a holomorphic function’s modulus at z = 0 in terms of its zeros.
|
|≤
|
|| |
35. Proof. This problem seems suspicious. For example, when z 0 = 0, of course nk=1 z0 zk = nk=1 zk > 1. When z0 is sufficiently large, it is also clear nk=1 z0 z k > 1. So the point z0 can be both inside and outside D(0, 1). I’m not sure what proof is needed for such a trivial claim.
| − | 23
| − |
| |
36. Proof. By Maximum Modulus Principle, M (r) = max|z|≤r f (z) . So M (r) is an increasing function on [0, R).
|
|
37. Proof. Suppose there is an n-th degree polynomial p(z) = a 0 + a1 z + + an z n such that n 1, a n = 0, and p(z) = 0 for any z C. Then q (z) = 1/p(z) is holomorphic over the whole complex plane. For any R > 0, we have by Maximum Modulus Principle
···
∈
max q (z)
|z |≤R
|
≥
| ≤ |max |q (z)|. z |=R
Since on z : z = R ,
{ | |
}
n 1
n
| p(z)| ≥ |an|R −|an−1|R −
−···−|a1|R−|a0| = R
n
|
|an−1 | − · · · − |a1| − |a0| → ∞ as R → ∞, an | − R Rn−1 Rn
we must have limR→∞ max|z|=R q (z) = 0. This implies supz∈C q (z) = 0, i.e. p(z) contradiction and our assumption must be incorrect.
|
|
|
|
≡ ∞.
This is a
38. Proof. Apply the Maximum Modulus Principle to the function g(z) = 1/f (z), which is holomorphic on U . 39. ∂ Proof. Note the Laplace operator ∆ = 4 ∂ 0 and log f (z) is a ¯ ∂ , so log z is a harmonic function on C ∂ harmonic function outside the zeros of f (z). Define K = z U : f (z) = 0 and V ε = z∈K D(z, ε). Then ¯ε and continuous on U ¯ V ε . By Maximum Modulus for any α R, α log z + log f (z) is harmonic in U V Principle for harmonic functions, max z∈U ¯ \V ε (α log z + log f (z) ) = maxz ∈∂ (U ¯ \V ε ) (α log z + log f (z) ). When z approaches to zeros of f (z), log f (z) , so by letting ε 0, we can further deduce that maxz∈U (α log z + log f (z) ) = maxz∈∂U (α log z + log f (z) ). Therefore
∈
||
||
|
|
|| { ∈ \ || | | | | → −∞ || | |
|
|
\{ } } ∪ \ || →
α log z + log f (z)
||
| ≤ max{α log r1 + log M (r1), α log r2 + log M (r2)}, ∀z ∈ U,
|
which is the same as α log r + log M (r)
≤ max{α log r1 + log M (r1), α log r2 + log M (r2)}, r ∈ [r1, r2].
Now let α be such that the two values inside the parentheses on the right are equal, that is α =
log M (r2 ) log r1
− log M (r1) . − log r2
Then from the previous inequality, we get log M (r)
≤ α log r1 + log M (r1) − α log r,
which upon substituting value for α gives log M (r) where s =
≤ (1 − s)log M (r1) + s log M (r2),
log r1 log r log r2 log r1 .
− −
40.
24
|
|
|
|
Proof. Fix r (0, 1). For any ρ (0, r) and any z integral formula, n, p N, we have
∈
∈
∀ ∈
n+ p
∈ D(0, ρ), we have zn ∈ D(0, ρ) (∀n ∈ N). By Cachy’s
n+ p
|
k
f (z )
k=n+1
|
n+ p
| − | − − | | · ≤ − − | | → k
=
f (z )
f (0) =
k=n+1 n+ p
≤
k=n+1
1 2π
2π
1 2π
≤
k =n+1
0
∂D (0,r)
∂D (0,r)
f (ζ ) dζ z k ζ (ζ z k )
f (reiθ ) dθ r ρ
−
1 2π
f (ζ ) ζ z k
n+ p
k =n+1
1 2π
f (ζ ) dζ ζ
2π
0
f (re iθ ) dθ k ρ r ρk
·
n+ p
ρk
0
k=n+1
∞ ¯ ρ). as n . So n=1 f (z n ) converges absolutely and uniformly on D(0, Remark 9. In general, there is no Mean Value Theorem for holomorphic functions. For example, f (z) = exp i 2z−z2(z−1z+1z2 ) π is analytic but f (z2 ) f (z1 ) = f (z)(z2 z1 ), z C (for more details, see Qazi [5]). But as the proof of Theorem 2.7 shows, the formula
→ ∞
−
f (z)
− f (z0) =
z
−
− z0
2πi
∂U
∀ ∈
f (ζ )dζ (ζ z)(ζ z0 )
−
−
more or less fills the void. In particular, it shows holomorphic functions are locally Lipschitz. 41. Proof. This problem is the same as Problem 15. 42. Proof. This problem is the same as Problem 19. 43. (1) Proof. By Problem 18,
≤ | | | | | ≤ ≤| | f (z ) 1 f (z )+1
−
z . So f (z) 1+ z 1 z
the above inequality implies f (z) From the same inequality
f (z ) 1 f (z )+1
−
|| −| | .
|− 1 ≤ |f (z) − 1| ≤ |z||f (z) + 1| ≤ |z||f (z)| + |z|. Since |z| < 1,
z , we have
− 1)2 + (Imf (z))2 ≤ |z|2[(Ref (z) + 1)2 + (Imf (z))2] ≤ |z|2(Ref (z) + 1)2 + (Imf (z))2. −|z| ≤ 1 < Ref (z). If So |Ref (z) − 1| ≤ |z |(Ref (z)+1). If Ref (z) > 1, we can deduce from this inequality 11+ |z| 1−|z | Ref (z) ≤ 1, we can deduce from this inequality 1 − Ref (z) ≤ |z |Ref (z) + |z |, i.e. 1+|z| ≤ Ref (z). (Ref (z)
(2) iθ
e z Proof. We note f (z) = 1+ if and only if 1−eiθ z Lemma.
f (z ) 1 f (z )+1
− = eiθ . So the claim is straightforward from Schwarz
44.
25
Proof. Because D(0, 1) is pre-compact, by Theorem 1.11 (Bolzano-Weierstrass Theorem), it suffices to find a sequence (zn )∞ D(0, 1), such that limn→∞ zn = z 0 ∂D(0, 1) and (f (zn ))∞ n=1 n=1 is bounded. Assume this does not hold. Then z0 ∂D(0, 1) and n N, there exists δ (z0 ) > 0 such that z D(z0 , δ (z0 )) D(0, 1), f (z) n. The family of these open sets (D(z, δ (z))z∈∂D (0,1) is an open covering of the compact set ∂D(0, 1). By Theorem 1.10 (Heine-Borel Theorem), there is a finite sub-covering ( D(zk , δ (zk )) pk=1 of ∂D(0, 1). Consequently, for each n N, we can find ε n > 0 such that z z : 1 εn z < 1 , f (z) n. Without loss of generality, we can assume (εn )∞ monotonically decreases to 0. n=1 Since f (z) uniformly approaches to as z ∂ D(0, 1), by Theorem 2.13, f has finitely many zeros in m D(0, 1). So f can be written as f (z) = i=1 (z zi )ki h(z), where h is a holomorphic function on ∂ D(0, 1) with no zeros in D(0, 1). So h satisfies the Minimum Modulus Principle: ρ > 0, h(z) min |ζ |=ρ h(ζ ) for any z D(0, ρ). Therefore, for any z D(0, 1 εn ),
⊂
|
|≥
∀ ∈
∈
∀ ∈
∀ ∈
∈
∀ ∈ {
− ≤ | |
∩
} |
|≥
∞
→ − · ∀ | | ≥ | | ∈ ∈ − |f (ζ )| ≥ n n |h(z)| ≥ |ζ |min | ≥ h(ζ )| = min . m m k |ζ |=1−ε | i=1(ζ − zi )k | max|ζ |=1−ε i=1 |ζ − zi |k =1−ε 2 Fix z ∈ D(0, 1 − εn ) and let n → ∞, we get h(z) = ∞. Contradiction.
n
n
i
n
m i=1
i
i
45. Proof. By assumption, we can write f (z) as f (z) = nj=1 (z z j )kj h(z), where h(z) is a holomorphic n function in D(0, 1) with no zeros in D(0, 1). Define G(z) = h(z) ¯ zj z)kj . Then G is holomorphic j =1 (1 on D(0, 1). Apply Maximum Modulus Principle to G(z) on z : z 1 ε (ε > 0), we get
n
| | −
|G(z)| ≤ |zmax |h(z) |=1−ε
1
j =1
¯ zj z
|k
j
= max
|z|=1−ε
z z
−
Since each 1−z¯jjz maps ∂ D(0, 1) to ∂ D(0, 1), by letting ε
− · · − { | | ≤ − } |f (z)| ≤ max k |z|=1−ε n z −z
j
j
j =1 1 z¯j z
−
1 n z zj j =1 1 z¯j z
− −
→ 0, we get G(z) ≤ 1, i.e.
46.
kj
.
| | ≤ f (z)
z zj n j =1 1 z¯j z
− −
kj
.
ζ +a Proof. Let a = f (0) and define ϕa (ζ ) = − 1−a ¯ζ . Then h(z) = ϕa (f (z)) maps D(0, 1) to D(0, 1) with h(0) = ϕa (a) = 0. So Schwarz Lemma implies h (0) 1. We note h (z) = ϕa (f (z)) f (z). Since
|
|≤ −(1 − a¯ζ ) − (−ζ + a)(−a¯) = −1 + |a|2 , ϕa (ζ ) = (1 − a ¯ζ )2 (1 − a ¯ζ )2
·
we have h (0) = ϕ a (a)f (0) = |af |2(0) −1 . So we have
|f (0)| ≤ 1, i.e. |f (0)| ≤ 1 − |a|2. 1 − |a|2 48. i + + Proof. The M¨obius transformation w(z) = zz− +i maps C to D = D(0, 1). So f Aut(C ) if and only if w f w−1 Aut(D). By Theorem 2.20, a D and τ R, such that w f w−1 (ζ ) = ϕ a ρτ (ζ ). Plain calculation shows
◦ ◦
∈
∃ ∈
f (z) = w −1 ϕa ρτ w(z) =
◦ ◦ ◦
∈
(1 + a)(z + i) (1 a)(z + i)
−
50. Proof. This problem is the same as Problem 10. 26
◦ ◦
∈
− (1 + ¯a)eiτ (z − i) . − (¯a − 1)eiτ (z − i)
◦
3
Theory of Series of Weierstrass
Throughout this chapter, all the integration paths will take the following convention on orientation: all the arcs take counterclockwise as their orientation and all the segments take the natural orientation of R1 as their orientation. 1. Proof. This is just Mittag-Leffler Theorem (Theorem 3.8). 2. Proof. To prove Theorem 3.1, choose varepsilon > 0 sufficiently small so that the circles z : z (1 k n) don’t intersect with each other or with ∂ U . Then by Cauchy’s integration theorem,
{ | − zk | = ε}
≤ ≤
n
f (z)dz
∂U
−
f (z)dz = 0,
|z−zk |=ε
k=1
which is ∂U f (z)dz = 2πi nk=1 Res(f, zk ). ¯ To prove Theorem 3.12, we choose R > 0 sufficiently large so that U
n
f (z)dz = 2πi
∂U
Since 3. (i)
∂D (0,R)
f (z)dz =
⊂ D(0, R). Then
Res(f, zk ) and
k=1
∞
−
f (z)dz = 0.
∂U
n k=1 Res(f, zk )
−2πiRes(f, ∞), we conclude
∈
f (z)dz
∂D (0,R)
Proof. Suppose z3 (z1+i) has Laurent series theorem, for ε (0, 1), cn
+ Res(f,
∞) = 0.
n −∞ cn (z + i) . Then by Theorem 3.2 and Cauchy integral
n=
1 1 1 1 dn+1 −3 dζ = (ζ ) 2πi |ζ +i|=ε (ζ + i)n+1 ζ 3 (ζ + i) (n + 1)! dζ n+1 ζ =−i 1 (n + 3)! (n + 2)(n + 3) n ( 1)n+1 ( i)−(n+4) = ( 1)n+1 i . (n + 1)! 2! 2
= =
·
−
−
−
(ii) Proof. We note z2 (z + 1)(z + 2)
=
(z + 1)(z + 2) 3z (z + 1)(z + 2)
− −2 = 1−
1 1 = 1+ z 1+
·
1 z
1 2 1+
− ·
z 2
=
3(z + 1) 1 1 = 1+ (z + 1)(z + 2) z + 1
−
∞ ( 1)k
− k=0
z k+1
−1
− z +4 2
∞ ( 1)k zk . 2k−1
− − k=1
(iii) Proof. We note log
− z z
−
a b
= =
− − − − − − − log
1 1
a z b z
∞ 1 a
n n=1
z
= log 1
n
+
∞ 1
n n=1
27
a z b z
log 1
n
=
∞ 1
n n=1
b z
(bn
an )z −n .
(iv) Proof. We note 2
z e
1 z
= z
∞ z−k
2
k =0
1 = z + z + + k! 2 2
∞
k=1
z −k . (k + 2)!
(v) 2k+1 ∞ ∞ 1)k z z 2k+1 z Proof. We note sin z = k=0 ( 1)k (2zk+1)! . So sin z+1 = k=0 (2(− k+1)! (z +1)2k+1 . Suppose sin z +1 has Laurent ∞ series n=−∞ cn (z + 1) n . Then by Theorem 3.2, for ε > 0,
cn
−
= = =
1 1 2πi |ζ +1|=ε (ζ + 1)n+1
∞
∞
k=0
( 1)k ζ 2k+1 dζ (2k + 1)! (ζ + 1) 2k+1
−
( 1)k 1 ζ 2k+1 dζ (2k + 1)! 2πi |ζ +1|=ε (ζ + 1) 2k+n+2 k=0
− k
≥
n−1 2
−
·
( 1)k 1 d2k+n+1 ζ 2k+1 (2k + 1)! (2k + n + 1)! dζ 2k+n+1
−
Since d2k+n+1 ζ 2k+1 dζ 2k+n+1 we have
− =
ζ = 1
0 (2k + 1)! (2k+1)! −n (−n)! ( 1)
cn
=
− −
=
−
0 sin1
1 ( n)! cos 1 1 ( n)! sin1
− −
Therefore z sin = z + 1
∞
k =0
n
( 1)k k 0 (2k+1)! ( 1)−n ( 1)k −n−1 k ( n)! (2k+n+1)! 2
≥
−
≥
n 1, n = 0, n 1,
≥
−
0
− .
ζ = 1
≤ −
≥ 1,
n = 0, n
≤ −1
n 1, n = 0, n 1, n is odd,
≥
≤ − n ≤ −1, n is even.
sin1 cos1 (z + 1) −2k + (z + 1) −(2k+1) . (2k)! (2k + 1)!
4. (i) Proof. z = 0 is a removable singularity. (ii) Proof. z = 1 is a pole of order 1. The Laurent series of the function in a neighborhood of z = obtained by
−1 can be
1 z2
−
πz cos 1 z + 1
By discussion (3) of page 91, z =
=
(z
−
1 1)(z + 1)
∞ ( 1)n
n=0
−1 is an essential singularity. 28
πz z + 1
2n
− (2n)!
.
(iii)
∞ z −n , we have z(e 1z − 1) = n=0
1
Proof. Since e z =
∞ z −n+1. Therefore z = 0 is an essential singularity. n=1
(iv) Proof. z = 1 is an essential singularity. (v) Proof. z = 1 is an essential singularity. z = 0 is a pole of order 1. (vi) Proof. z = kπ +
π 2
(k
∈ Z) are poles of order 1.
5. (1) Proof. The limit limz→a f (z) exists (finite or ) if and only if limz→a f (1z) exists. By discussion (3) on page 93, a is an essential singularity of f (z) if and only if a is an essential singularity of f (1z) .
∞
(2) Proof. Clearly, a is an isolated singularity of P (f (z)). We note by Fundamental Theorem of Algebra, P (ζ ) maps C to C. Consequently, P (ζ ) maps a dense subset of C to a dense subset of C. By Weierstrass Theorem (Theorem 3.3), f (z) maps a neighborhood of a to a dense subset of C. Therefore, P (f (z)) maps a neighborhood of a to a dense subset of C. So it is impossible for a to become a removable singularity or a pole of P (f (ζ )). Hence a must be also an essential singularity of P (f (z)). 6. (i)
∞
k Proof. We note any positive integer can be uniquely represented in the form =0 ak 2 , where each kn ∞ ak 0, 1 and only finitely many a k ’s are non-zero. In the expansion of n=0 (1 + z 2 ), each representation k z k=0 ak ·2 appears once and only once. Therefore, after a rearrangement of the terms, we must have
∈ { } ∞
∞
n
(1 + z 2 ) = 1 + z + z 2 + z 3 + z 4 +
n=0
·
··· = 1 −1 z .
(ii) Proof. sinh πz = 12 [eπz
− e−πz ] has zeros an = ni (n ∈ Z), where a0 = 0 is a zero of order 1. Since for any 2 ∞ ∞ R R = n=−∞,n n=−∞,n =0 |a | =0 n < ∞, by Weierstrass Factorization Theorem, we can
− 2
R > 0, we have n find an entire function h(z) such that
2
n=
∞
h(z )
sinh πz = ze
1
n=
−∞,n=0
Since limz→0
sinh πz z
z z e ni ni
h(z )
= ze
∞
z2 1+ 2 n
n=1
.
= π, we conclude eh(0) = π. Meanwhile we have π coth πz
= =
∞
1 z2 = ze h(z) 1+ 2 sinh πz sinh πz n n=1
(sinh πz)
∞
∞
2z 1 1 2z (z) + n2 + h (z) + = + h . 2 2 + z2 z z z n 1 + 2 n n=1 n=1
29
Let γ n be the rectangular path [n + (n + 12 )i, n + (n + 12 )i, n for any given a, when n is large enough, we have
− − (n + 12 )i, n − (n + 12 )i, n + (n + 12 )i]. Then
−
γ n
where I =
coth πz dz z 2 a2
2π[x+(n+ 1 )i] 2 +1 1 e2π[x+(n+ 2 )i] 1
e
−n
x + (n + 12 )i
n
= I + II + III + IV,
−
−
2
−
a2
II =
−n [x − (n + 12 )i]2 − a2 n+ 12
IV = We note
ex 1 ex +1
−
a2
dx,
−
−e2πx +1 −e2πx −1
n
−
2
−
−
−
2π[x−(n+ 1 )i] 2 +1 1 e2π[x−(n+ 2 )i] 1
x + (n + 12 )i
e2π(−n+yi) +1 e2π(−n+yi) 1 idy, n + yi)2 a2
e
n
and
n
(
n+ 12
III =
dx =
−(n+ 12 )
−e2πx +1 −e2πx −1
−n
dx =
−n [x − (n + 12 )i]2 − a2
dx
e2π(n+yi) +1 e2π(n+yi) 1 idy. (n + iy)2 a2
−
(n+ 12 )
−
−
∈ (−1, 1) for any x ∈ R. So we have n
1 −n x2 + (n + 12 )2
| |≤ I
−
2
dx = a2
(n +
1 2 2)
arctan
−
a2
n
(n +
1 2 2)
−
a2
→ 0, x
∈ R large enough, ee −+11 < 2. So
as n . Similarly we can conclude limn→∞ III = 0. We also note for x for n large enough,
→ ∞
1+e−2πn
n+ 12
| |≤ II
−(n+ 12 )
as n
1
≤ √ n24− a2 arctan √ nn2+−2a2 → 0,
e 1 e 2πn dy n2 + y 2 a2 −
x
−
−
→ ∞. Similarly we can conclude limn→∞ IV = 0. Combined, we have shown lim
coth πz dz = 0. z 2 a2
− πz Define f (z) = coth ∈ iZ, we have z −a . By Residue Theorem, for a n
→∞
2
γ n
2
n
f (z)dz = 2πiRes(f ; a) + 2πiRes(f ; a) + 2πi
−
γ n
Res(f ; an ).
k= n
−
πa It’s easy to see Res(f ; a) = Res(f ; a) = coth 2a . Since
−
lim (z
z
→an
− an)f (z) = zlim →a
n
we must have Res(f ; an ) = 1 0= lim 2πi n→∞
γ n
e2πz + 1 z an 2 2π(z an ) = lim 2πz = 2 2 2 2 2 πz z a e 1 2π(an a ) z→an e e2πan
−
−
−
−
−π(n 1+a ) . Therefore 2
2
f (z)dz = lim n
→∞
coth πa a
n
−
k= n
−
This implies for a i Z,
∈
−π(n2 1+ a2) ,
− −
1 π coth πa = + a 30
1 π(a2 + k2 )
∞
k=1
a2
2a . + n2
coth πa = a
∞
− k=
−∞
1 π(a2 + k 2 )
1 z
Plugging this back into the formula π coth πz = By Theorem 2.13, h (z) 0 for any z C. So
≡
∈
∞
+ h (z) +
z2 1+ 2 n
∞
2z n=1 n2 +z 2 ,
we conclude h (z) = 0 for z
∞
z2 n2
h(0)
sinh πz = ze
n=1
= πz
1+
n=1
∈ iZ.
.
Remark 10. The above solution is a variant of the proof for factorization formula of sin πz. See Conway[2] VII, 6 for more details.
§
(iii)
∞
π 2
1 4n2
− − − − − − − − − − − − − − − · − · − − · − · − − −
Proof. From problem (v), we have cos πz =
1 2
sin π
= π = π
1 2 1 2
= π
∞
1+
∞
z
1+
n=1
N
→∞ ∞
1 2
1
n=1
z n+
1
n=0
∞
1 2
1
1 2
z
1 2
1
n
1 2n
1
1
1 4n2
1 4n2
1
z
2
n2
n=1
n=1
N
= 1 and
z
N
z
1
π 2
lim
=
1 2
n=1
π = lim N →∞ 2 =
z
z
1
n=1
z
n
z n+
N
1
n=1
z N +
1 2n
1
1 2
z n+
1+
1 2
n
N
1 2
1+
n=1
N
1
1 2
z
n=1
z
n
1 2
z2
(n
1 2 2)
2
.
1 2
(iv) Proof. The function eπz
an = 2ni (n ∈ Z), where a0 = 0 is a zero of order 1. Since for any − 1 has zeros 2 ∞ R R = n=−∞,n =0 |a | =0 4n < ∞, by Weierstrass Factorization Theorem, we can n=−∞,n
−
∞
2
R > 0, we have n find an entire function h(z) such that πz
e
2
∞
h(z )
− 1 = ze
n=
1
−∞,n=0
z z e 2ni 2ni
h(z )
= ze
∞
n=1
z2 1+ 2 4n
.
To determine h(z), we note
− 1) = 1 + h(z) + ∞ 42nz = 1 + h(z) + ∞ 2z . −1 z eπz − z 4n2 + z 2 1 + 4zn n=1 n=1 Let γ n be the rectangular path [2n + (2n + 1)i, −2n + (2n + 1)i, −2n − (2n + 1)i, 2n − (2n + 1)i, 2n + (2n + 1)i]. πe πe πz
Define f (z) =
(eπz = πz 1 e
πz
(z 2
a2 )(eπz
−
2
2 2
−1) . Then for any given a, when n is large enough, we have
f (z)dz = I + II + III + IV,
γ n
31
where
−2n
I =
f (x + (2n + 1)i)dx =
2n
−(2n+1)
f ( 2n + yi)idy =
−
2n+1
III =
f (x
−2n
−(2n+1)
2n+1
IV =
f (2n + yi)idy =
2n
I
−2n
as n
2n+1
II
(4n2
−(2n+1)
+
y2
πdy a2 )(e2nπ
−
→ ∞. For n sufficiently large, e e
(2n + 1)2
− a2
|IV
|≤
−(2n+1)
2n
arctan
(2n + 1)2
2π 2n + 1 √ arctan √ ≤ → 0, nπ − 1) (e2 − 1) 4n2 − a2 4n2 − a2
πe 2πn dy (4n2 + y2 a2 )(e2πn
−
4π 2n + 1 √ arctan √ ≤ → 0, 2 2 − 1) 4n − a 4n2 − a2
lim
n
→∞
− a2 → 0
−1 ≤ 2, so
→ ∞. Combined, we have shown
By Residue Theorem, for a
− 1) idy.
2nπ
2nπ
2n+1
as n
−
− 1) idy,
− a2](eπx + 1) dx,
πe π (2n+yi) [(2n + yi)2 a2 ](eπ (2n+yi)
2π
− a2 ≤
πx
πe e (2n + 1)i)2
→ ∞. Similarly, we can show limn→∞ III = 0. Also, we note
| |≤ as n
−(2n+1)
πdx x2 + (2n + 1)2
| |≤
−
−2n [(x −
2n+1
−(2n+1)
It is easy to see
2n
− (2n + 1)i)dx =
− a2](eπx + 1) dx,
πe π(−2n+yi) [( 2n + yi)2 a2 ](eπ(−2n+yi)
−
2n+1
2n
and
πe e [(x + (2n + 1)i)2
2n
II =
πx
−2n
f (z)dz = 0.
γ n
∈ 2Zi, we have
n
f (z)dz = 2πiRes(f ; a) + 2πiRes(f ; a) + 2πi
−
γ n
It’s easy to see Res(f ; a) =
πe πa 2a(eπa 1) and
−
lim (z
z
→an
Res(f ; a) =
−
k= n
−
π 2a(eπa 1) .
−
Since
πe πz z an = 2 2 (z a ) eπz 1
− · − −
− an)f (z) = zlim →a
n
Res(f ; an ).
− 4n2 1+ a2 ,
−4n 1+a . Therefore
Res(f ; an ) =
2
2
1 0= lim 2πi n→∞ So
π eπa +1 2a eπa 1
− =
1 a2
+
∞
2 n=1 4n2 +a2
So h(z) = So
πz 2
π eπa + 1 2a eπa 1
n
−
1 4n2 + a2
− k=−n ∈ 2Zi. Therefore for z ∈ 2Zi, we have for a
πeπz h (z) = eπz
f (z)dz = lim
n
→∞
γ n
−1
1 z
∞
2z πeπz = 4n2 + z 2 eπz 1 n=1
πz
e
.
πz
π e + 1 π = . − πz − 2 e −1 2
− −
+ h(0). It’s easy to see lim z→0
eπz 1 z
− = π. So e h(0) = π. Combined, we conclude e h(z) = πe πz2 . ∞ πz z2
− 1 = πze
2
n=1
32
1+
4n2
.
Equivalently, we have z
e
− 1 = ze
∞
z 2
n=1
z2 1+ 2 2 4π n
.
(v) Proof. Using the formula sin πz =
−i sinh(iz) and the factorization formula for sinh πz, we have
sin πz =
−iπ(iz)
∞
z2 n2
∞
z2 n2
− − 1
n=1
= πz
1
n=1
.
7. For a clear presentation of convergence of infinite products, we refer to Conway [2], Chapter VII, 5. In particular, we quote the following theorem (Conway [2], Chapter VII, 5, Theorem 5.9)
§
§
Let G be an open subset of the complex plane. Denote by H (G) the collection of analytic functions on G, equipped with the topology determined by uniform convergence on compact subsets of G. Then H (G) is a complete metric space. Furthermore, we have the following theorem (proof omitted). Theorem 11. Let G be a region in C and let (f n )∞ n=1 be a sequence in H (G) such that no f n is identically
∞
zero. If [f n (z) 1] converges absolutely and uniformly on compact subsets of G, then n=1 f n (z) converges in H (G) to an analytic function f (z). If a is a zero of f then a is a zero of only a finite number of the functions f n , and the multiplicity of the zero of f at a is the sum of the multiplicities of the zeros of the functions f n at a.
−
∞
Proof. (Proof of Blaschke Product) Fix r (0, 1). Since k=1 (1 ak ) < r exists k 0 N, such that for any k k 0 , 1+ < a < 1. So for any k k 0 , k 2
− | | ∞, limk→∞ |ak | = 1. So there ≥ | | ≥ ak − z |ak | |ak |(1 − |ak |) + |z||ak |(1 − |ak |) ≤ (1 + r)(1 − |ak |) = 2 (1 − |ak |). 1 ≤ · − 1+r |1 − a¯k z||ak | 1−a ¯k z ak 1−r 2 (1 − r) ∈
∈
Since
∞ (1 − |a |) < k k=1
∞
· ∞
ak z k =1 1 a ¯k z
, we conclude
−
· |aakk | − 1 is absolutely and uniformly convergent on
− | ak | {z : |z| ≤ r}. Therefore − ak is absolutely and uniformly convergent on { z : | z | ≤ r }. By ∞ Weierstrass Theorem (Theorem 3.1), f (z) = k=1 1a−ka¯−kzz · |aakk | represents a non-zero holomorphic function on {z : |z| < 1}. Since the mapping z → 1a−ka¯−kzz (0 < |ak | < 1) maps D(0, 1) to D(0, 1), we conclude |f (z)| ≤ 1. By the theorem quoted at the beginning of the solution, it’s clear that (ak )∞ are the only zeros of f (z). ∞
ak z k=1 1 a ¯k z
−
k =1
8. Proof. Let R ε = R
− ε, ε ∈ (0, R). Define F ε (z) = f (z) 2
t Rε (z bj ) j =1 R2ε ¯ bj z s Rε (z ai ) i=1 R2ε a ¯i z
· 2
− − − −
, z
∀ ∈ D(0, Rε).
It’s easy to verify that each of RRε2(z−¯bjbzj ) and RRε2(z−a¯ai zi ) maps D(0, Rε ) onto itself and takes the boundary to ε ε the boundary. Therefore F ε (z) is analytic on D(0, Rε ), has no zeros in D(0, Rε ), and F ε (z) = f (z) for z = R ε . Since log F ε (z) is a harmonic function, by formula (2.33) on page 71,
||
|
1 log F ε (z) = 2π
|
|
−
−
|
|
2π
0
Rε eiφ + z 1 log F ε (Rε e ) Re dφ = iφ Rε e z 2π
|
iφ
|
−
33
2π
0
log f (Rε eiφ ) Re
|
|
| |
Rε eiφ + z dφ. Rε eiφ z
−
|
Plugging the formula of F ε into the above equality, we get 1 log f (z) = 2π
|
|
2π
0
Rε eiφ + z log f (Rε e ) Re dφ + Rε eiφ z iφ
|
|
−
t
Rε2 ¯bj z log Rε (z bj ) =1
j
Letting ε
→ 0 yields the desired formula.
− −
s
Rε2 a ¯i z log . R (z a ) ε i =1
− i
− −
9. Proof. By Theorem 3.4 (correction: “holomorphic function f (z)” in the theorem’s statement should be 1 2 3 7 “meromorphic function f (z)”), f (z) has the form of z+1 + z− 2 + (z −2)2 + c, where c is a constant. f (0) = 4 implies c = 1. Since for any ζ D(0, 1),
∈
1
1
− ζ )2 = 1 − ζ the Laurent expansion of f (z) in 1 < |z | < 2 is (1
f (z) = = = =
∞
∞
=
n
ζ
=
n=0
(n + 1)ζ n ,
n=0
1 2 3 + + +1 z + 1 z 2 (z 2)2 1 1 1 3 1 +1 z + 1 z 1+ z 1 2 4 (1 z2 )2
−
− − n 1 −z −
∞
1 z n=0
∞
− · − z 2
∞
3 z + (n + 1) 4 n=0 2
− − − ∞ ( 1)n zn
n=1
n=0
n
n
∞
+1
3 3n 1 n + + z . 4 n=1 2n+2
10. Proof. We follow the construction outlined in the proof of Mittag-Leffler Theorem (Theorem 3.9). For each n N, when z < n2 , ψ n (z) has Taylor expansion
∈
| |
ψn (z) =
n (z
−
1 = n)2 n
1
1 = n
∞
· − z 2 n
1
(k + 1)
k=0
z n
k
.
Let λn be a positive integer to be determined later. We estimate the tail error obtained by retaining only the first λ n terms of the Taylor expansion of ψ n (z):
ψn (z)
−
λn
1 n
z (k + 1) n
k=0
k
∞
≤
1 n
≤
1 1 + − λ 1 n n 2 n
=
1 1 + 1 − λ n 2 n n 2λn
k=λn +1
· ·
∞
k=λn
k +1
k=λn
·
∞ (k + 2)
2k+1
1 = n
∞
−
∞ ε < ∞ n=1 n
f (z) = U (z) +
∞
n=1
n
1 − 2 (z − n) n
where U (z) is an entire function. 34
+
1
2λn −1
∞ x
λn
log 12
d
1 2
x
λn 1 + . log 2 (log 2)2
1
2k+1
x 1 1 dx = + − x λ 1 n 2 n 2 n
·
k
k
k=λn
λn 1 + n·21λn log = n·21n 1 + n·12n 2 + (log 2)2 n·2λn 1 . By the proof of Mittag-Leffler Theorem, we can write f (z) as
If we let λn = n, then εn :=
(k + 1) 1 = k 2 n
n
(k + 1)
k =0
−
z n
k
,
n 1 log 2 + (log 2)2
satisfies
11. (i) 1 Proof. Define f (z) = (z2 −a2 )( C 2iZ. Let γ n be the rectangular path [2n+ (2n+ 1)i, 2n+ eπz −1) , where a (2n + 1)i, 2n (2n + 1)i, 2n (2n + 1)i, 2n + (2n + 1)i]. Then for any given a, when n is large enough, we have f (z)dz = I + II + III + IV,
− −
∈ \
−
−
γ n
where
−2n
I =
2n
f (x + (2n + 1)i)dx =
2n
II =
−2n
−(2n+1)
f (2n + yi)idy =
2n+1
−(2n+1)
III =
f (x
− − 1) , 2n −dx , −2n [(x − (2n + 1)i)2 − a2](eπx + 1) −
− (2n + 1)i)dx =
−2n
2n+1
2n+1
IV =
f (2n + iy)idy =
−(2n+1)
−(2n+1) [(2n
+ iy)2
It’s easy to see 2n
| |≤ I
−2n
as n
x2
dx + (2n + 1)2
− a2
−
arctan
→ ∞. Similarly, we can show limn→∞ III = 0. Also, we note 2n+1
| |≤ II
−(2n+1)
as n
2 (2n + 1)2
− ≤ a2
(4n2
+
y2
dy a2 )(1
−
− a2](eπx + 1) ,
idy a2 ](eπ(−2n+yi
[( 2n + yi)2
2n+1
2n
and
dx [(x + (2n + 1)i)2
idy a2 ](eπ(2n+iy)
2n (2n + 1)2
− 1) . − a2 → 0
2
2n + 1
− e−2nπ ) = (1 − e−2nπ )√ 4n2 − a2 arctan √ 4n2 − a2 → 0
→ ∞, and 2n+1
2n + 1 √ → 0 arctan 4n2 − a2 − 1) (e2πn − − a2 −(2n+1) (4n2 + y 2 − ∈ 2iZ, we have as n → ∞. Combined, we have shown limn→∞ γ f (z)dz = 0. By Residue Theorem, for a
|IV
dy a2 )(e2πn
|≤
2 1) 4n2
=
n
n
f (z)dz = 2πiRes(f ; a) + 2πiRes(f ; a) + 2πi
−
γ n
It’s easy to see Res(f ; a) =
z
we have Res(f ; 2ni) = 1 0= lim 2πi n→∞
γ n
1 2a(eπa 1) and
lim (z
→2ni
−
Res(f ; a) =
−
Res(f ; 2ni).
k= n
−
1 2a(1 e−πa ) .
−
Since
− 2ni) = − 1 , − 2ni)f (z) = z→lim2ni π(z2π(z − a2)(eπz − 1) π(4n2 + a2)
−π(4n1+a ) . Therefore 2
2
f (z)dz = lim n
→∞
1 e πa + 1 2a eπa 1
·
− −
1 π
n
k= n
−
1 2 4k + a2
1 e πa + 1 = 2a eπa 1
·
Replacing π a with z , we get after simplification 1 ez
−
1 = 1 z
This is the partial fraction of 1/(ez
−
1 + 2
∞
k=1
2z , z + z2
4k 2 π2
− 1). 35
∀ ∈ C \ 2Zπi.
− −
1 π
∞
k=
−∞
4k 2
1 . + a2
(ii) 1 Proof. Define f (z) = (z−a)sin C Z. Let γ n be the rectangular path [ n + 2 (πz ) , where a ni, n + 12 ni, n + 12 ni, n + 12 + ni]. Then when n is large enough, we have
− − −
1 2
∈ \
+ni,
−
n+
f (z)dz = I + II + III + IV,
γ n
where
−(n+ 12 )
−
I =
1
(n+ 2 ) dx = a)sin2 [π(x + ni)] −(n+ 12 ) (x + ni
4dx
− a)(e−2nπ+2ixπ + e2nπ−2ixπ − 2) , −n n −4idy idy II = = , 2 1 1 1 [ (n + 2 ) + yi − a]sin [−π(n + 2 ) + πyi] −n [−(n + 2 ) − a + yi](e2πy + e−2πy + 2) n (n+ ) (n+ ) dx −4dx III = = , 2 −(n+ ) (x − ni − a)sin [π(x − ni)] −(n+ ) (x − a − ni)(e2nπ+2iπx + e−2nπ−2iπx − 2) (x + ni
n+ 12
−
1 2
1 2
1 2
1 2
and n
IV =
1 2) +
−n [(n +
yi
−
n idy = a]sin2 [π(n + 12 ) + πyi] −n [(n + 12 )
−
4idy . a + yi](e−2yπ + e2yπ + 2)
It’s easy to see (n+ 12 )
− | |≤ I
−(n+ 12 )
as n
→ ∞,
(x
n
|II | ≤
−n n
≤ as n
0
1 2
(n +
+ 1) ≤ n(e2nπ4(2n − e−2nπ − 2) → 0
− e−2nπ − 2) n
=
+ a)2 + y2 (e2πy + e−2πy + 2)
(n +
0
8dy 1 2
+ a)2 + y2 (e2πy + e−2πy + 2)
− e−2πn → 0
8dy 8 1 = 1 1 πy 2 (n + 2 + a)e (n + 2 + a)
2π
→ ∞, (n+ 12 )
− | |≤ → ∞ − | | ≤ ≤ −(n+ 12 )
(x
4dx a)2 + n2 (e2nπ
4(2n + 1) ≤ − e−2nπ − 2) n(e2nπ − e−2nπ − 2) → 0
, and lastly n
IV
−n
(n +
n
0
as n
+ n2 (e2nπ
4dy
III
as n
4dx a)2
1 2
=
a)2 + y 2 (e2yπ + e−2yπ + 2)
8
(n +
n
4dy
8
− a)e2yπ = (n + 12 − a)
1
1 2
→ ∞. Combined, we have shown limn→∞
γ n
(n +
0
− e−2πn → 0
8dy 1 2
− a)2 + y2(e2yπ + e−2yπ + 2)
2π
f (z)dz = 0. By Residue Theorem, we have n
f (z)dz = 2πiRes(f ; a) + 2πi
γ n
k= n
−
36
Res(f ; k).
1 2
+
2
(z −k ) It’s easy to see Res(f ; a) = sin21(πa) . Define h k (z) = sin 2 (πz ) . Since limz →k h k (z) = a neighborhood of k and for ε > 0 small enough, and we have
Res(f ; k) =
1 2πi |z−k|=ε (z
We note d dz and
− hk (z) z a
hk (z) lim = z →k z a
lim
−
z
=
lim
z
=
→k
lim
z
=
→k
lim
z
= So for a
→k
→k
dz 1 hk (z) dz d = = 2πi |z−k|=ε (z a) (z k)2 dz a)sin2 (πz)
hk (z)(z = lim z →k (z
z =k
2(z
−
1 π 2 , h k is
− · −
holomorphic in
− hk (z) z a
. z =k
− a) − h(z) = lim hk (z) − 1 , z →k z − a − a)2 (k − a)2 π 2
− k)sin(πz) − 2π(z − k)2 cos(πz)
sin3 (πz) 2sin(πz) + 2π(z k)cos(πz) 4π(z k) cos(πz) + 2π 2 (z 3sin2 (πz) π cos(πz) 2sin(πz) 2π(z k)cos(πz) 3π sin2 (πz) 2π cos(πz) 2π cos(πz) + 2π2 (z k)sin(πz) 6π 2 sin(πz) cos(πz)
−
−
−
·
−
−
−
− k)2 sin(πz)
−
0.
∈ C \ Z,
2πi f (z)dz = sin2 (πa) γ n
n
−
Let n
→ ∞, we get
k= n
−
2πi . (k a)2 π 2
−
∞
π2 1 = . sin2 (πz) n=−∞ (z n)2
−
πz ) Remark 12. Another choice is to let f (z) = cot( (z +a)2 , see Conway [2], page 122, Exercise 6. The trick used in this problem and problem 6 (ii) will be generalized in problem 12.
(iii) Proof. By the solution of problem 6 (ii), we have
∞
1 2z coth z = + . 2 z n=1 z + π 2 n2
Since coth z = i cot(iz), we have
Therefore
∞
=
−
∞
− n=1
N
1 πn
− z −
− − − − − − − ··· − − − − − ··· − − − − − 1 β
1
αn
n=1
1 = lim N →∞ β =
2z 1 = π 2 n2 z 2 z n=1
−
cot z
π πβ cot = α α
∞
1 = i coth(iz) = z
lim
N
→∞ ∞
n=0
1 β
1 nα + β
β
1
α
β
1
α
β
1 1 = lim N →∞ β αn + β
+
1
2α
+
1 nα + (α
+ β 1 α + β
− β )
=
n=1
1
αn
β
1 πn + z
.
1 αn + β
1 1 1 1 + + + + + N α β α + β 2α + β N α + β 1 1 1 1 + + + 2α β (N 1)α + β Nα β N α + β
∞
α 2β (nα + β )[nα + (α n=0 37
− β )] .
···
−
−
By letting α = 3 and β = 1, we get π π π √ = cot = 3 3 3 3
∞
n=1
(3n
1 2)(3n
−
− 1) .
πz ) Remark 13. For a direct proof without using problem 6 (ii), we can let f (z) = cot( z 2 −a2 and apply Residue Theorem to it. For details, see Conway [2], page 122, Exercise 8. The line of reasoning used in this approach will be generalized in problem 12 below.
12. (1) Proof. Since z = is a zero of f (z) with multiplicity p, f (z) can be written as hz(pz) where h is holomorphic in a neighborhood of . Consequently, h is bounded in a neighborhood of . In the below, we shall work with a sequence of neighborhoods of that shrinks to . So without loss of generality, we can assume h is bounded and denote its bounded by M . Let γ n be the rectangular path [(n + 12 ) + ni, (n + 12 ) + ni, (n + 12 ) ni, (n + 12 ) ni, (n + 12 ) + ni]. Then f (z) cot(πz)dz = I + II + III + IV,
∞
∞
∞
∞
∞
−
−
−
−
γ n
where I =
eiπ(x+ni) +e−iπ(x+ni) 2 ni) p eiπ(x+ni) e−iπ(x+ni) 2i
−(n+ 12 ) h(x + ni)
(x +
(n+ 12 )
−
−n h(−(n + 1 ) + yi)
II =
2 1 2) +
=
[ (n +
and
as n
(n+ 12 )
n
II
as n
→ ∞, and
(n+ 12 )
−(n+ 12 )
|≤
− − −
p 2
1 2
−
eyπ e−yπ dy eyπ + e−yπ
−
−
−n (n + 12 )2 + y 2
2
2
p 2
−
eyπ e−yπ dy eyπ + e−yπ
−
38
→ 0
n ≤ n2M → 0 1 arctan + n+ 1
n + M enπ + e−nπ enπ + e−nπ 2M dx = arctan p − − nπ nπ nπ nπ e e e n n (x2 + n2 ) 2 e
M
− −−
−
n
|IV
iπ[−(n+ 1 )+yi] 2
−
n + M e−nπ + enπ e2nπ + 1 2M dx = arctan p e2nπ 1 n n (x2 + n2 ) 2 enπ e−nπ M
|≤
as n
− −
−n (n + 12 )2 + y 2
→ ∞, |III
iπ[−(n+ 1 )+yi] 2
h((n + 12 ) + yi) ( 1)n e−yπ i + ( 1)n eyπ ( i) ( 1)dy. −n [(n + 12 ) + yi] p ( 1)n e−yπ i ( 1)n eyπ ( i)
−(n+ 12 )
→ ∞,
1
1 2
| |≤ | |≤ I
1
eiπ[−(n+ 2 )+yi] +e−iπ[−(n+ 2 )+yi] 2
1 2
n
We note
2
idy − yi] p e −e 2i −n h(−(n + 1 ) + yi) (−1)n e−yπ (−i) + (−1)n eyπ i 2 (−1)dy, 1 [−(n + 2 ) + yi] p (−1)n e−yπ (−i) − (−1)n eyπ i n (n+ ) h(x − ni) eiπx+nπ + e−iπx−nπ III = idx, −(n+ ) (x − ni) p eiπx+nπ − e−iπx−nπ n
IV =
−(n+ 12 ) h(x + ni) eiπx−nπ + e−iπx+nπ dx = idx, (x + ni) p eiπx−nπ − e−iπx+nπ (n+ 1 )
1 2
→ 0
n ≤ n2M → 0 1 arctan + n+ 1 2
2
as n . Combined, we can conclude lim n→∞ by Residue Theorem, we have
→ ∞
f (z) cot(πz)dz = 2πi
γ n
γ n
f (z) cot(πz)dz = 0. Meanwhile, for n sufficiently large,
m
n
Res(f (z) cot(πz), αk ) + 2πi
k =1
Res(f (z) cot(πz), k).
k= n
−
We note 1 cos(πz) 1 ( 1)k f (z) cos(πz)(z k) f (k) Res(f (z) cot(πz), k) = f (z) dz = dz = . 2πi |z−k|=ε sin(πz) 2πi |z−k|=ε z k sin[π(z k)] π
So
−
f (z) cot(πz)dz = 2πi
γ n
−
m
n
Res(f (z) cot(πz), αk ) + 2πi
k =1
Letting n
→ ∞ gives us
k= n
−
n
f (k) . π
n
lim
n
− −
·
→∞ k −n =
f (k) =
− π
Res(f (z) cot(πz), αk ).
k=1
(2) Proof. As argued in the solution of (1), we can assume f (z) = hz(pz) where h is holomorphic in a neighborhood of and has bound M in that neighborhood. Let γ n denote the same rectangular path as in our solution of (1). Then f (z) dz = I + II + III + IV, γ n sin(πz)
∞
where I = II =
−(n+ 12 ) h(x + ni)
2idx , (x + ni) p eiπx−nπ e−iπx+nπ
n+ 12
−n h(−(n + 1 ) + yi)
[ (n +
−
n
III =
2 1 2) +
n+ 12
−(n+ 12 )
and
n
IV = We note
|I | ≤ as n
−
→ ∞, and |II | ≤
n
−n (n +
1 2) +
1
1
−n (n + 12 )2 + y 2
2i
idy, yi] p e−yπ (−1)n (−i) − eyπ (−1)n i
h(x ni) 2idx , p iπx nπ + (x ni) e e−iπx−nπ
− −
−
h((n + 12 ) + yi)
n+ 2 2M dx p − nπ nπ 2 e e −(n+ 12 ) (x + n2 ) 2
−
p 2
≤
2i
idy. yi e−πy (−1)n i − eπy (−1)n (−i) p
4M nπ e e−nπ
−
2M dy eyπ + e−yπ
n+ 12
n + dx 2M = arctan x2 + n2 (enπ e−nπ )n n
−
0
1 2
→ 0
n
Mdy 2M n = arctan n+ −n (n + 12 )2 + y 2 n + 12
≤
1 2
→ 0
as n . By similar argument, we can prove limn→∞ III = limn→∞ IV = 0. Combined, we can conclude f (z ) that limn→∞ γ n sin( πz ) dz = 0. Meanwhile, for n sufficiently large, by Residue Theorem, we have
→ ∞
γ n
f (z) dz = 2πi sin(πz)
m
Res
k =1
f (z) , αk sin(πz) 39
n
+ 2πi
Res
k= n
−
f (z) ,k . sin(πz)
It’s easy to see Res
f (z) ,k sin(πz)
1 ( 1)k = 2πi |z−k|=ε z k
Therefore,
γ n
f (z) dz = 2πi sin(πz)
Let n
→ ∞, we get limn→∞
− · f (z)(z − k) dz = (−1)k f (k) . π − sin[π(z − k)]
m
Res
k=1
n k= n (
−
n
( 1)k f (k) . π
− −
−1)k f (k) =
f (z) , αk sin(πz)
m k =1 Res
π
+ 2πi
k= n
−
f (z ) sin(πz ) , αk
.
(3) (i) Proof. Let f (z) =
1 (z +a)2 .
Then by result of (1),
∞
1 = 2 (n + a) n=−∞
−πRes
cot(πz) , a (z + a)2
−
2
−π dzd [cot(πz)]|z=−a = sin2π(πa) .
=
This result is verified by problem 11 (ii). (ii) Proof. Let f (z) =
∞
1 z 2 +a2 .
Then by result of (2),
( 1)n n2 + a2 n=−∞
−
=
−
=
π πcsch(πa) = − ai · sin(πai) . a
π Res
1 , ai + Res (z 2 + a2 )sin(πz)
−
1 , ai (z 2 + a2 )sin(πz)
This result can be verified by Mathematica. 13. (i) Proof. f (z) =
1 z 2 z 4 has
−1. ∞ is a removable singularity. 1 dz d 1 −(−2z) |z=0 = 0, Res(f ; 0) = = | z =0 = 2 2 2 2πi |z|=ρ z (1 − z ) dz 1 − z (1 − z 2 )2 1 dz 1 1 − | − Res(f ; 1) = = = , z =1 2πi (1 − z)(1 + z)z 2 (1 + z)z 2 2 −
poles 0, 1, and
and
|z−1|=ρ
1 dz Res(f ; 1) = 2πi |z+1|=ρ (1 + z)(1
−
−
z)z 2
=
1 . 2
(ii) Proof. We note f (z) :=
z 2 + z + 2 2 1+i 4 = + + z(z 2 + 1)2 z 4(z i)2 4(z
− − i + −1 − i + −4 + i . − i) 4(z + i)2 4(z + i)
−
− Therefore, the function f (z) has poles 0, i and −i. ∞ is a removable singularity of f (z). Furthermore, we have −4 − i = −1 − i , Res(f ; −i) = −4 + i = −1 + i . Res(f ; 0) = 2, Res(f ; i) = 4
4
40
4
4
(iii)
··· , an are the roots of the equation z n + an = 0. Then each ai is a pole of order 1 for z z +a . ∞ is a removable singularity of f (z). Then ani −1 an−1 (z − ai ) an−1 (z − ai ) 1 Res(f ; ai ) = n = lim in = lim i n = . n z →a z →a z +a n j =1,j j =1 (z − aj ) =i (ai − aj )
Proof. Suppose a1 , the function f (z) =
n−1
n
n
i
i
(iv) Proof. f (z) =
1 sin z
has π Z as poles of order 1.
∞ is not an isolated singularity of f (z). And 1 1 (−1)n (z − nπ) dz = dz = (−1)n . |z−nπ|=ε sin z 2πi |z−nπ|=ε z − nπ sin(z − nπ)
1 Res(f,nπ) = 2πi
(v) Proof. First, we note as cos ( 1)n z 3 n=0 (z 2)2n .
∞
− −
1
=
z 2
−
1 2
ez
i
2
−
+ e− z
i
( 1)n n=0 (z 2)2n .
=
2
−
∞
− −
So f (z) := z 3 cos
This shows z = 2 is an essential singularity of f (z), and
Res(f ; 2) =
∞ ( 1)n
−
n=0
2πi
|z−2|=ε
1 z 2
−
=
z3 dz = ( 1) 3z 2 (z 2)2n
− · |z=2 + (−1)2 · 3!1 · 6|z=2 = −11.
−
Also,
∞ is an isolated singularity of f (z), and we have Res(f ;
∞)
= = =
−
− − − | −
1 z 3 cos 2πi |z|=R
1
z
1 1 +2 2πi |ζ |= 1R ζ 1 d4 (ζ + 2) 3 cos ζ 4! dζ 4
−
2
dz =
3
cos ζ
ζ =0 =
1 z 3 cos 2πi |z−2|=R
1
z
−2
dz
dζ 1 (ζ + 2) 3 = cos ζdζ ζ 2 2πi |ζ =ε| ζ 5
8 . 3
(vi) Proof. Let f (z) = 1
ez z (z +1)
= ez
− e−1.
− 1 z
1 z +1
. Then Res(f ;0) = 1, Res(f ; 1) =
−
−e−1, and Res(f ; ∞) =
14. Proof. We can write g(z) as (z
− a)2h(z) where h is holomorphic and h(a) = 0. Then h(a) = (zg−(aa)) g (a)(z − a) − 2g(a) h (a) = . (z − a)3
41
2
and
Therefore
Res
f (z) ,a g(z)
1
f (z) · 2 (z − a) h(z) f (a) (zg−(aa)) − f (a) g (a)((zz−−aa))−2g(a)
= Res =
f (z) d ,a = h(z) dz
2
f (a)h(a) f (a)h (a) h2 (a)
−
= z =a
3
g2 (a)
(z a )4
=
−
− a)[f (a)(z − a) + 2f (a)] − f (a)g(a)(z − a)2 .
g(a)(z
g 2 (a)
15. We make an observation of some simple rules that facilitate the evaluation of integrals via Residue Theorem. Rule 1 (rule for integrand function). The poles of the integrand function should be easy to find, such that the integrand function can be easily represented as (zf −(za))n , where f is holomorphic. The holomorphic function f (z) can be multi-valued function like logarithm function and power function. When it’s difficult to make f holomorphic in the desired region, check if it is the real or imaginary part of a holomorphic function. Rule 2 (rule for integration path). The choice of integration path is highly dependent on the properties of the integrand function, where symmetry and multi-valued functions (log and power functions) are often helpful. Oftentimes, the upper and lower limits of integration either form a full circle (i.e. [0, 2π]) so that substitution for trigonometric functions can be easily done or have as one of the end points.
∞
(i) 2
{ −R ≤ Rez ≤ R, Imz = 0}, γ 2 = {z : |z| = R, 0 ≤ arg z ≤ π }, and f (z) = (z z+1) . Then
Proof. Let γ 1 = z :
2
2
d z2 π f (z)dz + f (z)dz = 2πi Res(f, i) = 2πi lim = , z →i dz (z + i)2 2 γ 1 γ 2
and
π
f (z)dz =
0
γ 2
as R
·
·
R2 e2θi Reiθ idθ θi 2 2 2 (R e + 1)
·
dx → ∞. So by letting R → ∞, we have 0∞ (xx +1) 2
(ii)
2
2
=
1 2
π
≤ 0
R3
− 1)2 dθ → 0
(R2
∞ x2 dx π −∞ (x2 +1)2 = 4 .
x Proof. We note sin 2 x = 1−cos2 , so 2 π 2
0
1 dx = 2 2 a + sin x
π
0
1 dx = 2 2 a + sin x
2π
0
dθ . (2a + 1) cos θ
−
Let b = 2a + 1, then π 2
0
dx 1 = 2 2 |z|=1 b a + sin x
The equation 12 z 2 + bz have z 1 D(0, 1) and z 2
−
∈
π 2
0
dz iz
−
− 12 = 0 has two roots: ¯ 1). Therefore, ∈ D(0,
dx = i a + sin2 x |z|=1 (z
−
dz z1 )(z
1 = 1 −1 ) 2i |z|=1 2 (z + z
z1 = b
− z2 )
− √ b2 − 1 and z2
= i 2πi
·
42
−
dz + bz
− 12 . √ = b + b2 − 1.
1 2 2z
· z1 −1 z2 = √ b2π− 1 = 2
Since b > 1, we
π . a(a + 1)
Remark 14. If R(x, y) is a rational function of two variables x and y , for z = e iθ , we have R(sin θ, cos θ) = R Therefore
2π
− − 1 2
1 z
z +
R(sin θ, cos θ)dθ =
0
,
1 2i
z
1 z
1 1 1 (z + ), (z 2 z 2i
R
|z|=1
dz . iz
, dθ =
1 ) z
dz . iz
Remark 15. When the integrand function is a rational function of trigonometric functions, in view of the previous remark, it is desirable to have [0, 2π] as the integration interval. For this reason, we first use symmetry to expand the integration interval from [0, π2 ] to [0, π], and then use double angle formula to expand [0, π] to [0, 2π]. This solution is motivated by Rule 2. (iii) iz
{ −R ≤ Rez ≤ R, Imz = 0}, γ 2 = {z : |z| = R, 0 ≤ arg z ≤ π }, and f (z) = zze+1 . Then
Proof. Let γ 1 = z :
2
f (z)dz +
γ 1
f (z)dz = 2πiRes(f, i) =
|z−i|=ε
γ 2
and
π
f (z)dz =
γ 2
as R
0
iθ
ReiRe Reiθ idθ 2 2 θi R e +1
·
ze iz dz ie−1 π = 2πi = i, (z i)(z + i) 2i e
·
−
≤
π
0
R2 e−R sin θ dθ R2 1
→ 0
−
→ ∞ by Lebesgue’s Dominated Convergence Theorem. Therefore by letting R → ∞, we have ∞ xeix ∞ x cos x ∞ x sin x π dx = dx + i dx = i, 2 2 2 e −∞ x + 1 −∞ x + 1 −∞ x + 1
which implies (iv)
∞ x sin x dx =
0
x2 +1
π 2e .
Proof. Let r, R be two positive numbers such that r < 1 < R. Let γ 1 = z : r z R, arg z = 0 , γ 2 = z : r z R, arg z = π , γ R = z : z = R, 0 arg z π , and γ r = z : z = r, 0 arg z π . log z Define f (z) = (1+ ) and takes the principle branch of Logz with z 2 )2 where log z is defined on C [0, arg z (0, 2π) (see page 23). By Residue Theorem,
{ ≤ | | ≤ ∈
}
{ | |
≤
\ ∞
{
≤ }
≤ | | ≤ { | | ≤ 2
d log z f (z)dz = 2πiRes(f, i) = 2πi lim = z →i dz (z + i)2 γ 1 +γ R +γ 2 −γ r
} ≤ }
−π2 + π4 i.
We note
→ ∞
π
f (z)dz =
0
γ R
as R
,
π
f (z)dz =
γ r
as r
→ 0, and
0
·
log(reiθ ) re iθ idθ (r2 e2iθ + 1)2
f (z)dz =
γ 2
π
≤ ≤ − ≤ ≤ − −
log(Reiθ ) Reiθ idθ (R2 e2iθ + 1)2
−r
−R
·
log x dx = (1 + x2 )2
0
π
0
R
r
43
R (log R)2 + θ 2 dθ (R2 1)2
r
(log r)2 + θ 2 dθ (1 r2 )2
log( x) dx = (1 + x2 )2
R
r
πR(log R + π) (R2 1)2
−
πr( log r + π) (1 r2 )2
−
−
log x + πi dx. (1 + x2 )2
→ 0
→ 0
Therefore by letting r
→ 0 and R → ∞, we have ∞ 2log x + πi
(1 + x2 )2
0
i.e. (v)
∞
0
log x (1+x2 )2 dx =
dx =
−
π π2 + i, 2 4
−π4 . 1−α
z 1−α = e (1−α)log z is defined Proof. Let γ 1 , γ 2 , γ r and γ R be defined as in (iv). Define f (z) = 1+ z 2 , where z on C [0, ) with log z taking the principle branch of Logz. By Residue Theorem,
\ ∞
f (z)dz = 2πiRes(f, i) = 2πi
γ 1 +γ R +γ 2 γ r
−
We note
f (z)dz =
γ R
as R
→ ∞,
→ 0, and
R1−α eiθ(1−α) Reiθ idθ iθ 2 2 1+R e
·
0
π
f (z)dz =
γ r
as r
π
0
·
r 1−α eiθ(1−α) reiθ idθ 1 + r2 e2iθ
·
−r x1−α f (z)dz = dx = −R 1 + x2 γ 2
·
e (1−α) log z z + i
R
r
·
|z=i = πe (1−α)
R2−α π R2 1
π 2i
.
≤ → − ≤ → −
0
r2−α π 1 r2
e(1−α) log(−x) dx = 1 + x2
R
r
0
x1−α ei(1−α)π dx. 1 + x2
Therefore by letting r
→ 0 and R → ∞, we have
∞ x1−α (1 + ei(1−α)π )
0
i.e. (vi)
∞ x1 α dx = 1+x2
−
0
π 2
csc
1+
x2
απ 2
.
dx =
∞ x1−α
0
1+
x2
dx 2cos
·
(1
− α)π ei(1−α) 2
π 2
π
= πei(1−α) 2 ,
Proof. This problem is a special of problem (x). See the solution there. (vii) Proof. We use the method outlined in the remark of Problem (ii). π
0
dz dθ 1 π dθ 1 iz = = a + cos θ 2 −π a + cos θ 2 |z|=1 a + z +2z
The equation 12 z 2 + az + 12 = 0 has two roots: z1 = z1 = a+√ 1a2 −1 < 1 and z2 > 1. So
| |
| |
π
0
dθ 1 = a + cos θ i |z|=1 (z
−
dz z1 )(z
1
−
1 dz = 1 2 2i |z|=1 2 z + az +
−a + √ a2 − 1 and z2
− z2)
= 2π
=
1 2
.
−a − √ a2 − 1.
· z1 −1 z2 = √ a2π− 1 .
Remark 16. The expansion of integration interval from [0, π] to [ π, π] is motivated by Rule 2.
−
(viii)
44
Clearly,
∞ sin x dx = π , we have x 2
−
Proof. Using the result on Dirichlet integral:
∞ sin x
2
dx =
x
0
∞
2
sin xd
0
0
1 x
∞ 2sin x cos xdx
=
x
0
∞ sin(2x)
=
2x
0
d(2x) =
π . 2
(ix) Proof. It is easy to see that when λ = p = 0, the integral is equal to 1. So without loss of generality, we only consider the cases where λ and p are not simultaneously equal to 0. Choose r, R so that 0 < r < R, and r is sufficiently small and R is sufficiently large. Define γ 1 = z : r z R, arg z = 0 , γ 2 = z : r z R, arg z = 2π , γ r = z : z = r, 0 < arg z < 2π , and γ R = z : p p log z z = R, 0 < arg z < 2π . Let ρ = cos λ, then ρ ( 1, 1]. Finally, define f (z) = z2 +2zzcos λ+1 = z2e+2ρz +1 . For R sufficiently large and r sufficiently small, the two roots of z 2 + 2rz + 1 = 0 are contained in z : r < z < R . These two roots are, respectively, z1 = ρ + i 1 r2 = cos λ + i sin λ and z2 = ρ i 1 r2 = cos λ i sin λ . So z1 = z2 = 1, and arg z1 +arg z2 = 2π. Denote arg z1 by θ. If λ = 0, we have by Residue Theorem
≤ | | ≤ || { √ | | −− −
}
} −
{
}
−|
≤ | | ≤
|
}
∈ −
}
√ −
−
| | | |
f (z)dz
{ | |
−
|
e p log z1 e p log z2 2πi[Res(f, z1 ) + Res(f, z2 )] = 2πi + z1 z2 z2 z1 π 2πi [e pθi e p(2π−θ)i ] = sin(λp)e pπi . sin λ sin λ
=
γ 1 +γ R γ 2 γ r
− −
=
|
−
|
{ {
−
−
|
−
If λ = 0, we have by Residue Theorem
f (z)dz = 2πiRes(f, 1) =
−
γ 1 +γ R γ 2 γ r
− −
−2 pπi · e pπi .
Meanwhile, we have the estimates
f (z)dz
γ R
as R
→ ∞,
f (z)dz
γ r
as r
→ 0, and
f (z)dz =
γ 2
Since 1
≤ ≤
R
− 0
R2
1
R p 2R
− − 1 · R · 2π → 0
−
r p 2ρr
− r2 · r · 2π → 0
(xe2πi ) p dx = x2 + 2ρx + 1
R
− 0
x p dx e2 pπi . x2 + 2ρx + 1
− e2 pπi = 2 sin2( pπ) − 2sin( pπ) cos( pπ)i = −2i sin( pπ)e pπi , by letting R → ∞ and r → 0, we have π sin(λp) ∞ if λ =0 x p dx sin λ sin( pπ ) = 2 x + 2x cos λ + 1 pπ csc( pπ) if λ = 0 and p = 0. 0
If we take the convention that λp) single one: sinπ λsin( sin( pπ ) .
α sin α =
1 for α = 0, then the above three formulas can be unified into a
Remark 17. We choose the above integration path in order to take advantage of the multi-valued function x p (Rule 1). The no symmetry in x 2 + 2x cos λ + 1 is handled by using the full circle instead of half circle. (x)
45
Proof. Choose two positive numbers r and R such that 0 < r < 1 < R. Let γ 1 = z : r z R, arg z = 0 , γ 2 = z : r z R, arg z = 2π , γ R = z : z = R, 0 < arg z < 2π and γ r = z : z = r, 0 < arg z < 2π . log z 1 1/p Define f (z) = p(zz+1)z where z 1/p = e p is defined on C [0, ). Note by substituting y p for x, we get
{
≤ | | ≤
}
{ | |
1 + x p
0
}
\ ∞
∞ dx
∞
=
0
{ ≤ | | ≤ { | |
} }
1
y p dy . p(y + 1)y
By Residue Theorem, 1
( 1) p f (z)dz = 2πRes(f, 1) = 2πi = p γ 1 +γ R −γ 2 −γ r
π . p
where α =
−
− 2π
f (z)dz =
→ ∞,
0
f (z)dz =
→ 0, and
0
·
iθ
1
e p log(re ) reiθ idθ p(reiθ + 1)reiθ
·
R
f (z)dz =
γ 2
iθ
1
e p log(Re ) Reiθ idθ p(Reiθ + 1)Reiθ
2π
γ r
as r
−2πi e p
log eπi p
−2iαeαi,
=
We have the estimates
γ R
as R
· −−
r
1
R
(xe2πi ) p dx = p(x + 1)x
− r
≤
≤
1
2πR p p(R 1)
− → 0 1
2πr p p(1 r)
− → 0
1
x p dx e2αi . p(x + 1)x
·
Therefore by letting r
→ 0 and R → ∞, we have ∞ dx ∞ x dx −2iαeαi = −2iαeαi = α = π csc = = 1 + x p 1 − e2αi −2sin αe( +α)i sin α p 0 0 p(x + 1)x
1 p
π 2
π p
.
Remark 18. The roots of 1 + x p = 0 are not very expressible. So following Rule 1, we make the change of variable x p = y. Then we choose the above integration path to take advantage of the multi-valued function 1 y p . Since there is no symmetry in the denominator (y + 1)y, we used a full circle instead of a half circle. The change of variable x p = y is equivalent to integration on an arc of angle 2π/p instead of 2π. (xi) Proof. (Oops! looks like my solution has a bug. Catch it if you can.) By the change of variable x = have ∞ 1 1− p x (1 x) p y p dy dx = . 1 + x2 (y + 1) 3 + (y + 1) 0 0
−
2
Res(f, zi ) =
i=0
=
1 (z + 1)(z
= 0.
− z1) z=z
2
1 1 + + i ( 2i) i 2i
−·−
we
The equation (y + 1)3 + (y + 1) = 0 has three roots: z0 = p f (z) = (z+1)3z+(z+1) . Then
1 y +1 ,
·
−1, z1 = −1 + i, and z2 = −1 − i.
1 + (z + 1)(z 1 i i
−·
46
− z2) z=z
+ 1
(z
−
1 z1 )(z
− z2) z=z
0
Define
Let R and r be two positive numbers with R > r. Define γ 1 = z : r < z R, arg z = 0 , γ 2 = z : r < z R, arg z = 2π , γ R = z : z = R, 0 < arg z < 2π , and γ r = z : z = r, 0 < arg z < 2π . Then by Residue Theorem
| |≤
}
{ | |
}
f (z)dz = 2πi
γ 1 +γ R γ 2 γ r
− −
f (z)dz =
ΓR
as R
2π
→ ∞, and
2π
∞
(re iθ ) p reiθ idθ (reiθ + 1)3 + (reiθ + 1)
·
y p dy + 0 (y + 1) 3 + (y + 1)
0
}
{
Res(f, zi ) = 0.
(Reiθ ) p Reiθ idθ 3 iθ iθ (Re + 1) + (Re + 1)
→ 0. So by letting r → 0 and R → ∞, we have
·
0
0
as r
}
i=0
We note
| | ≤ { | |
2
{
∞
− 0
≤
2
≤
R3
−
2πr p+1 4r 3r2
− −
2πR p+1 3R2 4R
− − 2 → 0
− r3 → 0
y p dy e2 pπi 3 (y + 1) + (y + 1)
·
− 0 = 0.
Therefore..., “Houston, we got a problem here.” (xii) Proof. Choose r, R such that 0 < r < R and r is sufficiently small and R is sufficiently large. Let γ 1 = z :r z R, arg z = 0 , γ 2 = z : r z R, arg z = π , γ R = z : z = R, 0 < arg z < π and log z γ r = z : z = r, 0 < arg z < π . Define f (z) = z 2 +2z+2 . By Residue Theorem,
{
≤ | | ≤ { | |
}
{
}
≤ | | ≤
·
−
2π
f (z)dz =
γ R
as R
→ ∞,
→ 0, and
0
·
2π
log(reiθ ) reiθ idθ r2 e2iθ + 2reiθ + 2
·
0
R
f (z)dz =
γ 2
Therefore by letting r
−
log(Reiθ ) Reiθ idθ iθ iθ 2 2 R e + 2Re + 2
f (z)dz =
γ r
as r
r
log( x) dx = x2 2x + 2
−
−
→ 0 and R → ∞, we have
∞
0
∞ i.e. 0
log x log x + 2 dx + i 2 x 2x + 2 x + 2x + 2
−
−
+
log x x2 +2x+2
R
0
≤ ≤
x2
}
log2 3 + πi . 2 4
log R + 2π 2πR R2 2R 2
− −
log r + 2π 2πr 2 2r r2
− −
→ 0
→ 0
log x + πi dx. x2 2x + 2
−
r
∞
log x x2 2x+2
{ | |
f (z)dz = 2πi Res(f, 1 + i) = π
γ 1 +γ R +γ 2 γ r
We note
}
πdx π 3 = log2 + π 2 i, 2x + 2 2 4
−
dx = π2 log 2.
∞ log x − log x dx = ∞ log x ∞ log y 4x Meanwhile, we note 0 x2 − 2x+2 (x2 +2)2 −4x2 dx = 0 y 2 +4 dy. To calculate x2 +2x+2 0 ∞ y dy, we apply again Residue Theorem. Let’s define g (z) = log z and then we have the value of 0 ylog 2 +4 z 2 +4 g(z)dz = 2πi Res(g, 2i) =
·
γ 1 +γ R +γ 2 γ r
We note
−
g(z)dz =
γ R
2π
0
log(Reiθ ) Reiθ idθ R2 e2iθ + 4
·
47
≤
π π log 2 + i . 2 2
log R + 2π 2πR R2 4
−
→ 0
as R
→ ∞,
g(z)dz =
γ r
as r
→ 0, and
2π
·
0
R
g(z)dz =
γ 2
So by letting r
∞
0
−
r
∞ log x
x2 + 4
0
Hence
log( x) dx = x2 + 4
≤ R
r
log r + 2π 2πr 4 r2
−
→ 0
log x + πi dx. x2 + 4
→ 0 and R → ∞, we have 2
log(re iθ ) iθ re idθ r2 e2iθ + 4
log x π x2 +4 dx = 4 log 2.
dx + πi
∞ dx
π π 2 = log2 + i. x2 + 4 2 4
0
Solving the equation
∞ log x dx + ∞ log x dx = π log2 2 2x+2 2 0 x2 +2x+2 ∞x− ∞ log x log x π log2, 0 x2 −2x+2 dx − 0 x2 +2x+2 dx = 4
0
we get
∞
x2
0
log x π dx = log 2. + 2x + 2 8
Remark 19. Note how we handled the combined difficulty caused by no symmetry in the denominator of the integrand function and log function: because x2 + 2x + 2 is not symmetric, we want to use full circle; but this will cause the integrals produced by different integration paths to cancel with each other. Therefore we are forced to choose half circle and use two equations to solve for the desired integral. (xiii) Proof. We choose r and R such that 0 < r < R, and r is sufficiently small and R is sufficiently large. Let γ 1 = z : r z R, arg z = 0 , γ 2 = z : r √ z R, arg z = π , γ R = z : z = R, 0 < arg z < π , and z γ r = z : z = r, 0 < arg z < π . Define f (z) = zz2log +1 . Then
{ ≤ | | ≤ { | |
}
}
{
≤ | | ≤
}
f (z)dz = 2πiRes(f, i) =
γ 1 +γ R +γ 2 γ r
−
We note
√ −
2π
f (z)dz =
γ R
as R
→ ∞,
γ r
as r
→ 0, and
R
f (z)dz =
γ 2
r
By letting r
0
f (z)dz =
0
x log( x) dx = x2 + 1
−
·
1
(re iθ ) 2 log(reiθ ) iθ re idθ r2 e2iθ + 1
·
R
0
∞ √ x log x dx = π√ 2 . x2 +1 0 2 2
√ −
≤ ≤
r
√
2πR R(log R + 2π) R2 1
−
π
x2 + 1
dx
−π
∞
0
48
√ x x2 + 1
√
2πr r(log r + 2π) 1 r2
π xe 2 i (log x + πi) dx = e 2 i 2 x +1
→ 0 and R → ∞, we have ∞ √ x log x (1 + i)
Therefore,
1
√
}
π2 π i π2 ie 4 = ( 1 + i). 2 2 2
(Reiθ ) 2 log(Reiθ ) iθ Re idθ R2 e2iθ + 1
2π
{ | |
dx =
−
√ R
r
→ 0
→ 0
π x log x dx + e 2 i πi 2 x +1
π2 ( 1 + i). 2 2
√ −
R
r
√ x x2 + 1
dx.
(xiv) n
n
∞ z and ln(1 + z) = ∞ ( 1)n+1 z hold in Proof. We note the power series expansion ln(1 z) = n=1 n n=1 n z < 1 and the convergence is uniform on z 1 ε, for any ε (0, 1). Therefore, for any δ > 0, we have
||
| | ≤
∞ ∞
− − ∞
x
log
δ
e +1 ex 1
dx =
δ
Therefore
∞
n+1
( 1)
e−nx
log
0
ex + 1 ex 1
−
+
n
n=1
−
− −
∞ e−nx n
n=1
dx = lim 2 δ
→0
∈
dx = 2
k=1
(2k
∞ ∞ e−(2k−1)x
δ
∞ e−(2k−1)δ
−
=2
1)2
2k
k=1
∞
k =1
−1
−
dx = 2
∞ e−(2k−1)δ
k=1
(2k
− 1)2 .
1
(2k
− 1)2 ,
where the last equality can be seen as an example of Lebesgue’s Dominated Convergence Theorem applied ∞ π2 1 to counting measure. By Problem 11 (ii) with z = 12 , we know k=1 (2k− = 2 1) 8 . So the integral is equal
π2 4 .
to
(xv) Proof. Since we have argued rather rigorously in (xiv), we will argue non-rigorously in this problem.
∞
x dx = ex + 1
0
∞ e−x x dx = 1 + e−x 0
∞
e−x x
0
∞
−
n
( 1) e−nx dx =
n=0
∞
∞
− n
( 1)
e−(n+1)x xdx =
0
n=0
n=0
π Using the function π2 sin πz and imitating what we did in Problem 12, we can easily get 2 ∞ x Therefore 0 ex +1 dx = π12 .
(xvi)
∞ (−1)n
∞
n=1
(n + 1)2
( 1)n−1 n2
−
=
.
π2 12 .
Proof. We provide two solutions, which are essentially the same one. Solution 1. In problem (xvii), we shall prove (let a = 1) π
0
Note
π
0
So
π 2
0
x sin x dx = 2 2cos x
−
log(sin θ)dθ =
π
0
x sin x dx = π log2. 2 2cos x
−
x 2sin x2 cos x2 dx =2 2 2sin2 x2
·
·
π 2
0
θ cos θdθ = sin θ
π 2
− 2
log(sin θ)dθ.
0
−π2 log.
Solution 2 . We provide a heuristic proof which discloses the essence of the calculation. Note how Rule 1-3 lead us to this solution. π 2
log(sin θ)dθ
=
0
π 2
0
π
= =
π
1 2 log(cos α)dα = log(cos α)dα 2 − π2
1 2 1 + cos 2α log dα 4 − π2 2 1 π π log(1 + cos β )dβ log2. 8 −π 4
−
(Note how Rule 2 leads us to extended the integration interval from [0 , π2 ] to [ π, π].) Motivated by the similar difficulty explained in the remark of Problem (xvii), we try Rule 1 and note
−
π
−π
iβ
log(1 + e )dβ =
|z|=1
log(1 + z) dz = 2π log 1 = 0. iz
49
For the above application of Cauchy’s integral formula to be rigorous, we need to take a branch of logarithm function that is defined on C ( , 0] and take a small bypass around 0 for the integration contour, and then take limit. Using the above result, we have
\ −∞
π
1 π 1 π 0 = Re log(1 + e )dβ = log (1 + cos β )2 + sin2 β dβ = [log 2 + log(1 + cos β )] dβ. 2 −π 2 −π −π π
iβ
Therefore −π log(1 + cos β )dβ = π 2
0
−2π log 2 and
1 π π log(sin θ)dθ = log(1 + cos β )dβ log 2 = 8 −π 4
−π2 log2.
−
(xvii) Proof. We first assume a > 1. Then by integration-by-parts formula we have π
0
1
−
1 π x sin xdx 2 −π 1 2a cos x + a2 1 π xd(1 2a cos x + a2 ) 2 −π (1 2a cos x + a2 ) 2a π 1 2 2π log(1 + a) log(1 4a −π
x sin x dx = 2a cos x + a2
= =
−
−
−
· −
2
− 2a cos x + a )dx
.
Since a > 1, a ζ z : Rez > 0 for any ζ ∂ D(0, 1). Therefore we can take a branch of the logarithm ¯ 1). For example, function such that log(a z) is a holomorphic function on D(0, 1) and is continuous on D(0, iπ we can take the branch log z such that it is defined on C ( , 0] with log(e ) = π and log(e−iπ ) = π. By Cauchy’s integral formula,
− ∈ {
}
−
∈
\ −∞
π
log(a
−π
π
iθ
−e
)dθ =
|z|=1
−
log(a z) dz = 2π log a. iz
−
π
Therefore, −π log(1 2a cos x + a 2 )dx = 2Re −π log(a a calculation of the original integral, we get πa log 1+ . a If 0 < a < 1, we have
−
π
0
1
−
x sin x 1 dx = 2a cos x + a2 a2
π
0
1
−
x sin x 2 a cos x +
− eiθ )dθ = 4π log a.
1 dx = 1 a2 a2
π
· 1 log a
1+
Plug this equality into the
1 a
1 a
=
π log(a + 1). a
Assuming the integral as a function of a is continuous at 1, we can conclude for a = 1, the integral is equal to π log 2. The result agrees with that of Mathematica. To prove the continuity rigorously, we split π π xdx x sin xdx π the integral into 02 1−2xasin cos x+a2 and π 1−2a cos x+a2 . For the second integral, we have (x [ 2 , π])
∈
2
1
−
x sin x x sin x = 2 2a cos x + a a eix
x sin x ≤ | − | |a − ei | ≤ x sin x. π 2
So by Lebesgue’s dominated convergence theorem, the second integral is a continuous function of a for a (0, ). For the first integral, we note (1 2a cos x + a2 ) takes its minimum at cos x. So
∈ ∞
−
x sin x x . 2 1 2a cos x + a sin x Again by Lebesgue’s dominated convergence theorem, the first integral is a continuous function of a for a (0, ). Combined, we conclude the integral
≤
−
∈ ∞
π
0
as a function of a
1
−
x sin x dx 2a cos x + a2
∈ (0, ∞) is a continuous function. 50
1
−
Remark 20. We did not take the transform cos x = z +2z because log(1 2a cos x + a2 ) would then become 1 log[w(a) w(z)] log(2a), where w(z) = z +2z maps D(0, 1) to U := C [ 1, 1] and we cannot find a holomorphic branch of logarithm function on U . The application of integration-by-parts formula and the extension of integration interval from [0, π] to [ π, π] are motivated by Rule 2. But here a new trick emerged: if the integrand function or part of it could come from the real or imaginary part of a holomorphic function, apply Cauchy’s integral formula (or integration theorem) to that holomorphic function and try to relate the result to our original integral (Rule 1).
−
−
−
−
\−
−
(xviii) Proof. The function log(z + i) is well-defined on C+ := z : Imz 0 . We take the branch of log(z + i) that evaluates to log 2 + π2 i at i. Define γ R = z : z = R, 0 arg z π . Then for R > 1, Residue Theorem implies R log(z + i) log(z + i) i + i π2 dz + dz = 2πi = π log 2 + i. 1 + z2 i+i 2 −R 1 + z2 γ R
{ ≤
{ | |
Note
γ R
as R
→ ∞. So
≥ } ≤ }
log(z + i) dz = 1 + z2
π
0
·
log(Reiθ + 1) iθ Re idθ 1 + R2 e2iθ
·
≤
πR[log(R + 1) + 2π] R2 1
∞ log(x+i) π2 −∞ 1+x2 dx = π log 2 + 2 i. Hence ∞ log(x + i) ∞ log √ x2 + 1 π log 2 = Re dx = dx = −∞ 1 + x2 −∞ 1 + x2
−
∞ log(1 + x2 )
0
1 + x2
→ 0 dx.
Remark 21. The main difficulty of this problem is that the poles of 1+1x2 coincide with the branch points of log(x2 + 1), so that we cannot apply Residue Theorem directly. An attempt to avoid this difficulty x2 +1) x+i) x−i) might be writing log( as the sum of log( and log( and integrate them separately along different 1+x2 1+x2 1+x2 paths which do not contain their respective branch points. But log(x + i) and log(x i) cannot be defined simultaneously in a region of a Riemann surface which contains both integration paths. But the observation that Re[log(x i)] = log x2 + 1 gives us a simple solution (Rule 1).
−
√
±
16.
∞ z n is divergent on every point of ∂ D(0, 1): if z = 1, n=0
Proof. The power series if z = 1 and z = e iθ , N
N
∞ z n = 1+1+1 + ··· ; n=0
N
θ sin (N +1) cos N2θ sin ( N +2 1)θ sin N2θ 2 z = cos(nθ) + i sin(nθ) = +i . θ θ sin sin 2 2 n=1 n=0 n=0 n
∞
Similarly, the series n=0 z −n−2 is divergent on every point of ∂D(0, 1). So the functions represented by these two power series cannot be analytic continuation of each other. 17.
∞
Proof. It’s clear we need the assumption that α = 0. The series n=0 (αz)n is convergent in U 1 = z : z < n ∞ 1 1 n [(1−α)z ] |α| . The series 1−z n=0( 1) (1−z)n is convergent in U 2 = z : 1 α z < 1 z . On U 3 = U 1 U 2 , both of the series represent the analytic function 1−1αz . So the functions represented by these two series are analytic continuation of each other.
}
−
{ | − || | | − |}
18.
51
{ | | ∩
Proof. On the line segment z : 0 < Rez < 1, Imz = 0 , Taylor series of ln(1 + z) is z 12 z 2 + 13 z 3 z 1 On the same line segment, 0 < 1− x) ( x 1, x = 1) gives 2 < 2 . So Taylor series of ln(1
{
}
−
z (1 − z)2 (1 − z)3 − ln 2 − − 2 · 22 − 3 · 23 − · · · = ln 2 + ln 2 1
−
| | ≤ 1−z 1− = ln(1 + z).
2
−···.
Since the two holomorphic functions represented by these two series agree on a line segment, they must agree on the intersection of their respective domains (Theorem 2.13). Therefore they are analytic continuations of each other. 19. Proof. Clearly the series is convergent in D(0, 1) and is divergent for z = 1. So its radius of convergent n ∞ R = 1. Let f (z) = n=0 z 2 . Then f (z) is analytic in D(0, 1) and z = 1 is a singularity of f (z). We note n n 1 n 1 ∞ ∞ ∞ f (z) = z + n=1 z 2 = z + n=1 z 2·2 = z + n=1 (z 2 )2 = z + f (z 2 ). Therefore, we have
−
−
f (z) = z + f (z 2 ) = z + z 2 + f (z 4 ) = z + z 2 + z 4 + f (z 8 ) = n
··· .
So the roots of equations z 2 = 1, z 4 = 1, z 8 = 1, , z 2 = 1, , etc. are all singularities of f . These roots form a dense subset of ∂D(0, 1), so f (z) can not be analytically continued to the outside of its circle of convergence D(0, 1).
···
4
· ··
Riemann Mapping Theorem
3. Proof. Let g(z) = ∞ cn z n . Then g(z) is holomorphic and has the same radius of convergence as f (z). n=0 ¯ z∗ R convergence domain of g (z), we have
∀ ∈ ∩
g(z∗ ) =
∞
n=0
∞
n
c¯n z∗ =
n=0
¯ z∗ ) = f (z ¯ ∗ ) = f (z∗ ), c¯n (z¯∗ )n = f (¯
where the next to last equality is due to the fact that z∗ R and the last equality is due to the fact that f (R) g in the intersection of their respective domains of convergence. So R. By Theorem 2.13, f cn = c¯n , i.e. cn (n = 0, 1, ) are real numbers.
⊂
···
∈
≡
11. ¯ z ) (first take conjugate of z, then take conjugate of f (¯z )). Then Proof. Let g (z) = f (¯ ∂g(z) ∂ ¯ f (¯ z) ∂f (¯z ) = = conjugate ∂ ¯ z ∂ ¯ z ∂z
= 0.
¯ U ) ¯ = So g is a holomorphic function. Clearly g is univalent. Since U is symmetric w.r.t. real axis, g(U ) = f ( ¯ ¯ z0 ) = f (z ¯ 0 ) = 0, and f (U ) = conjugate(D(0, 1)) = D(0, 1). Furthermore, g (z0 ) = f (¯ g (z0 ) =
lim
z
→z0
¯ z) f (¯ z
¯ z0 ) − f (¯ − z0
f (¯z ) f (¯ z0 ) = lim conjugate z →z0 z¯ ¯ z0 = conjugate(f (¯ z0 )) = conjugate(f (z0 )) = f (z0 ) > 0. By the uniqueness of Riemann Mapping Theorem, g
≡ f . 52
− −
