Complex Numbers
What is Complex Number? • Complex Number is a combination combination (sum or difference) of real and imaginary numbers.
Definition of Complex Numbers • A complex number, number, z, is a number in the form of a + b i or x + jy, where i or j j = -1. • a or x is called the real part of z, and b or y is calle called d the the imagin imaginary ary part of z. z. • The The stan standa dard rd form form of a comp comple lex x num numbe berr is, z = a + bi = x + jy, where a, b, x and y are real numbers.
Definition of Complex Numbers • It can be represented geometrically either as points, or as directed line segments (vectors), in the complex plane.
Definition of Complex Numbers Complex numbers as points.
Definition of Complex Numbers Complex numbers as vectors.
Complex Plane • Known as Argand Diagram in honor to Swiss amateur mathematician and bookkeeper Jean-Robert Argand, which introduced the concept of the complex plane in 1806. • It is also called the z-plane because of the representation of complex numbers in the form z = x + iy.
Complex Plane • Plotting the complex number z = a + ib = x + jy as the point (a, b) or (x, y) in the plane (rectangular cartesian coordinates) where: x-coordinate of z is a = Re{z} y-coordinate is b = Im{z}. • A complex number written in the form z = a + ib is said to be expressed in cartesian form or rectangular form.
Forms of Complex Numbers • Rectangular Form: z = (x ± jy) where: x = real part or component jy = imaginary part or imaginary component.
Forms of Complex Numbers • Polar Form:
z r
where: r = magnitude or amplitude θ = argument or displacement in degrees.
= bar angle
Forms of Complex Numbers • Trigonometric Form:
z r (cos j sin ) rcjs rcis
Forms of Complex Numbers • Exponential Form:
z re
j
where: r = magnitude or amplitude θ = argument or displacement in radians.
Example Convert the following: a. 6 j 3 to polar form, exponential form and trigonometric form. O 6 30 b. to rectangular form, exponential form and trigonometric form.
Solution a. polar form r
6 3 2
2
45 6.708
3 tan 26.56 6 z r 6.708 26.56 1
Solution a. exponential form r
6 2 32
45 6.708
3 tan 0.4636 26.56 180 6 j j 0.4636 z re 6.708e 1
Solution a. trigonometric form r
6 3 2
2
45 6.708
3 tan 26.56 6 z r (cos j sin ) z 6.708(cos(26.56) j sin( 26.56)) z 6.708(cos 26.56 j sin 26.56) 1
Solution b. rectangular form x r cos 6 cos(30) 5.196 y r sin 6 sin( 30) 3 z x jy 5.196 j 3
Solution b. exponential form r 6
30 z re
j
180
6e
0.524 j 0.524
Solution b. trigonometic form r 6
30 z r cos j sin z 6cos(30) j sin( 30) z 6cos 30 j sin 30
Imaginary Number • It is a real number with an imaginary operator either i or j. where: i or j = pure imaginary unit =
1
Integral Powers of i or j i i,
i i i i,
i 1,
i i i 1, i
5
6
2
4
4
2
i i i i, i i i i, 3
7
2
4
3
i i i 1, i i i 1, 4
2
2
8
4
i
4
i i
4 n 1
i,
4n2
1,
4 n 3
i ,
4n4
1,
Example Find the equivalent of the following: a. i1995 b. i2006 c. i1988 d. i1991
Solution Applying trial and error a. 1995 = 4n+1
1995 = 4n+4 n = (1995-1)/4 = 498.5 n = (1995-4)/4 n = 497.75 1995 = 4n+2 n = (1995-2)/4 = 498.25 1995 = 4n+3 n = (1995-3)/4 = 498
i1995 = - i
Solution Applying trial and error b. 2006 = 4n+1
2006 = 4n+4 n = (2006 -1)/4 = 501.25 n = (2006 - 4)/4 n = 500.5 2006 = 4n+2 n = (2006 - 2)/4 = 501 2006 = 4n+3 n = (2006 - 3)/4 = 500.75
i2006 = - 1
Solution Applying trial and error c. 1998 = 4n+1
1998 = 4n+4 n = (1998 -1)/4 = 499.25 n = (1998 - 4)/4 n = 498.5 1998 = 4n+2 n = (1998 - 2)/4 = 499 1998 = 4n+3 n = (1998 - 3)/4 = 498.75
i1998 = - 1
Solution Applying trial and error d. 1991 = 4n+1
1991 = 4n+4 n = (1991-1)/4 = 497.5 n = (1991 - 4)/4 n = 496.75 1991 = 4n+2 n = (1991 - 2)/4 = 497.25 1991 = 4n+3 n = (1991-3)/4 = 497
i1991 = - i
Theorems on Complex Numbers 1. If x + iy = 0, then x = 0, and y = 0. 2. If x1 + iy1 = x2 + iy2, then x1 = x2 and y1 = y2. 3. If (x1 + iy1)(x2 + iy2) = 0, then at least one of the factors is zero, that is , x 1 + iy1 = 0 or x2 + iy2 = 0.
Arithmetic Operations in Rectangular Form a. Addition: z1 + z2 = (x1 + iy1)+(x2 + iy2) = (x1 + x2)+i(y1 + y2). b. Subtraction: z1 - z2 = (x1 + iy1)-(x2 + iy2) = (x1 - x2)+i(y1 - y2).
Arithmetic Operations in Rectangular Form c. Multiplication: z1z2 = (x1 + iy1)(x2 + iy2) = (x1x2 - y1y2 ) + i(x1y2 + x2y1). d. Division: z1 z2 z1 z2
x1 iy1 x2 iy2
x1 iy1 x2 iy2
x1 x2 y1 y2 x y 2 2
2 2
i
x2 iy2 x2 iy2
x2 y1 x1 y2 x y 2 2
2 2
Arithmetic Operations in Rectangular Form e. Extraction of Square Roots:
x jy
a jb
or 1
1
x jy 2 r ( 360 k ) 2
0
2
Arithmetic Operations in Polar Form and Exponential Form • The representation of z by its real and imaginary parts is useful for addition and subtraction. • For multiplication and division, representation by the polar form and exponential form has apparent geometric meaning.
Multiplication of two Polar Forms of Complex Numbers z1 r 1 cos 1 j sin 1 r 1 1 , z2 r 2 cos 2 j sin 2 r 2 2 . z1 z2 r 1 1 r 2 2 z1 z2 r 1r 2 cos 1 j sin 1 cos 2 j sin 2 z1 z2 r 1r 2 [(cos 1 cos 2 sin 1 sin 2 ) j (sin 1 cos 2 cos 1 sin 2 )] z1 z2 r 1r 2 [cos( 1 2 ) j sin( 1 2 )] z1 z2 r 1r 2 1 2
Multiplication of two Exponential Forms of Complex Numbers z1 r 1e
j 1
z 2 r 2 e
j 2
z1 z 2 r 1e
j 1
z1 z 2 r 1r 2 e
r e
j 2
2
j ( 1 ( 2 ))
Division of two Polar Forms of Complex Numbers z1
r 1 1
cos 1 j sin 1 cos 2 j sin 2 r 2 cos 2 j sin 2 cos 2 j sin 2 r 1
z2
r 2 2
z1
r 1 cos 1 cos 2 sin 1 sin 2
z2
r 2
z1
r 1
z2
r 2
z1
r 1
z2
r 2
2
cos
2
sin
2
2
j
cos( 1 2 ) j sin( 1 2 ) 1 2
sin 1 cos 2 cos 1 sin 2 2
cos
2
sin
2
2
Division of two Exponential Forms of Complex Numbers z1 r 1e
j 1
z2 r 2e z1 z2
j 2
r 1e
j 1
r 2e
j 2
r 1 r 2
e
j ( 1 2 )
Example Perform the indicated operations: a. 6 j 7 10cjs 30 b. 5 j 3 630 O c.
5 j 4 3 j 4
O
10e
j 0.752
5 25
O
Solution a.
6 j 7 10cjs30 10e 5 25 6 j 7 1030 1043.09 5 25 6 j 7 8.66 j5 7.303 6.831 4.532 j 2.113 6 8.66 7.303 4.532 j 7 5 6.831 2.113 9.175 j 6.718 O
j 0.752
O
Solution b. rectangular form
5 j35.196 j3 2 (5)(5.196) j (3)(3) ( j3)(5.196) (5)( j3) 25.98 9 j15.588 15 16.98 j30.588 34.9860.96
Solution b. polar form
5.83130.96630 (5.831)(6)30.96 30 34.98660.96
Solution c.
5 j 4 3 j 4
15 16 j12 j 20 9 16 j12 j12 3 j 4 3 j 4
31 j8 25
1.24 j 0.32
Example • Evaluate the square root of (3+j4).
Solution x jy 3 j 4
x jy
3 j 4
2
x j 2 xy j y 3 j 4 2
2
2
x y j 2 xy 3 j 4 2
2
real : x y 3 2
2
eqn.1
imaginary : 2 xy 4; y
4 2 x
2 x
Solution 2
2 x 3 x 2 4 2 3 x x 2 x 2
x x
2 2
4 3 x
2 2
3 x 4 0 2
2
Solution Using quadratic formula:
x 2
3 9 16 2
() x 2
x
35 2
4 2
8 2
4
3 25 2
35 2
Solution Using quadratic formula:
( ) x 2
x
35 2
2 2
1
1 imaginary(drop )
x 2
and y
2 x
2
2
1
Solution 2 j 3 j 4 first root : 2 j sec ond root : 2 j
Solution (alternative) 360 k x jy 2 r 2 3 j 4 553.13 first root : k 0 1
1
2
53.13 360 (0) 5 2 2.236 26.56 2 j 1
2
Solution (alternative) sec ond root : k 1
53.13 360 (1) 5 2 2.236 206 .56 2 j 1
2
Seatwork 1.1 1. Find the sum and difference of 5cjs 30 O 2e j 0.752 3 j 5
2. Simplify using rectangular form and polar form 4 j3 2 j 3. Find the square root of the product of 3 j 4 and 2 j8
Powers of Complex Numbers and De Moivre’s Theorem j r Let z re 2 2 j 2 2 r 2 ; successive powers, z r e z r e
r 3 ,... n r n ,
3 j 3
3
In general,
z r e n
n
jn
3
letting r = 1,
cos j sin
n
cos n j sin n
or in the abbreviated form.
n
n
Example 2 2 1. Simplify cos 2 j sin 2 cos 6 j sin 6 3 cos 3 j sin 3 cjs10 3 2. Evaluate 1 3 2 j 2
Solution 1. cos 2 j sin 2 2 cos 6 j sin 6 2 3 cos3 j sin 3 cjs10 cos j sin 4 cos j sin 12 9 10 cos j sin cos j sin
cos j sin 19 cos j sin 16
1
cos j sin
3
1
cos3 j sin 3
Solution 3
2. 1 3 j 2 2 1 j 3 cos j sin 2 2 cos 1 ; cos1 1 60 2 2 sin 3
2
; sin
1
3
2
60
cos 60 j sin 60 cos180 j sin 180 3
1
Roots of Complex Numbers Let
z re
j ( 2 k )
r k 360
0
Then 1 n
2 k j n n
1
z z n r e 1
r n
k 360 n
0
, k 0,1,2,...
Roots of Complex Numbers 0 0 k k 360 360 j sin r n cos n n 0 0 360 k 360 k n j sin W k r cos n n 1
where: k = 0,1,2,…(n-1) W0 = is the principal value or root, and θ must be a positive angle.
Example 1. Find the three roots of 125 2. Find the four roots of 16 20
0
Solution 1.
125 1250 first root : k 0 0 0 0 360 ( 0 ) 0 360 ( 0 ) j sin W 0 3 125 cos 3 3 W 0 5cos 0 j sin 0 5cjs 0 50
Solution 1.
sec ond root : k 1 0 0 0 360 (1) 0 360 (1) 3 j sin W 1 125 cos 3 3 W 1 5cos120 j sin 120 5cjs120 5120
Solution 1.
third root : k 2 0 0 0 360 (2) 0 360 (2) 3 j sin W 2 125 cos 3 3 W 2 5cos 240 j sin 240 5cjs 240 5240
Solution 2.
16 20
0
first root : k 0
20 360 (0) 20 360 (0) W 0 16 cos j sin 4 4 W 0 2cos(5) j sin( 5) 2cjs(5) 2 5 2355 4
Solution 2.
sec ond root : k 1
20 360(1) 20 360(1) W 1 16 cos j sin 4 4 W 1 2cos85 j sin 85 2cjs85 285 4
Solution 2.
third root : k 2
20 360(2) 20 360(2) W 2 16 cos j sin 4 4 W 2 2cos175 j sin 175 2cjs175 2175 4
Solution 2.
fourth root : k 3
20 360(3) 20 360(3) W 3 16 cos j sin 4 4 W 3 2cos 265 j sin 265 2cjs 265 2265 4
Homework 1.1 1. Find the three roots of 1230
2. Simplify 2 cos
3
j sin
0
3
6
Exponential and Trigonometric Functions of Complex Numbers From Euler’s formulas
e cos z j sin z jz e cos z j sin z z z1 z 2 jz
putting
e e
j ( z1 z 2 )
cos( z1 z 2 ) j sin( z1 z 2 )
j ( z1 z 2 )
cos( z1 z 2 ) j sin( z1 z 2 )
Exponential and Trigonometric Functions of Complex Numbers By addition and subtraction,
sin( z1 z 2 ) cos( z1 z2 )
e
j ( z1 z 2 )
e
j ( z1 z 2 )
j 2 e
j ( z1 z 2 )
e 2
Eqn. 1 j ( z1 z 2 )
Exponential and Trigonometric Functions of Complex Numbers e e e
j ( z1 z 2 )
e
j ( z1 z 2 )
cos z1 j sin z1 cos z2 j sin z2
j ( z1 z 2 )
cos z1 cos z2 sin z1 sin z2
jz1
e
jz 2
j sin z1 cos z2 cos z1 sin z2 Eqn. 2
Exponential and Trigonometric Functions of Complex Numbers e e e
j ( z1 z 2 )
jz1
e
j ( z1 z 2 )
cos z1 j sin z1 cos z2 j sin z2
j ( z1 z 2 )
cos z1 cos z2 sin z1 sin z2 j sin z1 cos z2 cos z1 sin z2
e
jz 2
Eqn. 3
Exponential and Trigonometric Functions of Complex Numbers Substituting Eqn. 2 and Eqn.3 to Eqn. 1,
sin( z1 z 2 ) sin z1 cos z 2 cos z1 sin z 2 , cos( z1 z 2 ) cos z1 cos z 2 sin z1 sin z 2
jy Then z1 z 2 x jy Let z1 x, z 2
Exponential and Trigonometric Functions of Complex Numbers sin( x jy ) sin x cos jy cos x sin jy y
sin( x jy ) sin x
e
2
e
y
cos x
e
y
sin( x jy ) sin x cosh y j cos x sinh y cos( x jy ) cos x cosh y j sin x sinh y sin( jy ) j sinh y cos( jy ) cosh y
j 2
y
e
Hyperbolic Functions of Complex Numbers hyperbolic functions for the complex number z
sinh z
e e z
2
z
, cosh z
e e z
2
z
,
Hyperbolic Functions of Complex Numbers cosh z sinh z 1 2
2
sinh( z1 z2 ) sinh z1 cosh z 2 cosh z1 sinh z 2 cosh( z1 z 2 ) cosh z1 cosh z2 sinh z1 sinh z 2 sinh( x jy ) sinh x cos y j cosh x sin y cosh( x jy ) cosh x cos y j sinh x sin y sinh( jy ) j sin y cosh( jy ) cos y
Example Determine the value of each of the following: a.
sin( j 0.78)
c.
sinh( 0.942 j 0.429)
d.
cos(0.942 j 0.429)
b.
cosh( j 0.78)
Solution a.
sin( j 0.78) j sinh( 0.78) j 0.861 b.
cosh( j 0.78) cos(0.78) 0.711
Solution c.
sinh( 0.942 j 0.429 )
sinh 0.942 cos 0.429 j cosh 0.942 sin 0.429 0.989 j 0.615 d.
cos(0.942 j 0.429 )
cos 0.942 cosh 0.429 j sin 0.942 sinh 0.429 0.643 j 0.358
Seatwork 1.2 Determine the value of each of the following: a. tan( j 0.78) b. tan( 0.942 j 0.429 ) 2 2 sin ( 0 . 942 0 . 429 ) cos (0.942 j 0.429) 1 j c.
Logarithms of Complex Numbers Express the complex number z = x + jy in the general exponential form
z re
j ( 2 k )
where θ is in radians and k = 0, ±1, ±2,… Taking the natural logarithms of both numbers,
ln z ln( x jy ) ln r j ( 2k ), where : k 0,1,2,...
Two Types Types of Logarithm: Logarithm: 1. Common(or Brigssian) Logarithm Notation: log Base: 10 ; i.e. log10Z 2. Natural(or Napierian) Logarithm Notation: ln Base: e = 2.718281828… ; i.e. logeZ = lnz
Properties of Logarithm: 1. logbN = x ; N = bx 2. logeN = y ; ln N = y ; N = ey 3. lnex = x 4. elny = y 5. 10logx = x 6. lnxn = nlnx 7. loga(xy) = logax + logay 8. loga(x/y) = logax - logay
Logarithm and Natural Logarithm of a Complex Numbers
Lo L og ( A ) Log A e
j / 18 180 0
Lo L og ( A ) LogA Lo L og e
180 0 j / 18
Lo L og ( A ) LogA j / 180 Lo L og e Lo L og ( A ) LogA j / 180 0.4343
Logarithm and Natural Logarithm of a Complex Numbers Ln( A ) Ln A e
j / 180
Ln( A ) LnA Lne
j / 180
Ln( A ) LnA j / 180 Lne Ln( A ) LnA j / 180 1
Logarithm and Natural Logarithm of a Complex Numbers Log ( N ) Log N 180 Log ( N ) LogN j180 / 180 Loge Log ( N ) LogN j 0.4343
Logarithm and Natural Logarithm of a Complex Numbers Ln( N ) Ln N 180 Ln( N ) LnN j180 / 180 Lne Ln( N ) LnN j 1
Example Determine the general value of the following : a. ln 630 b.(3 j 2) c. Log(-9) d. Ln(-9)
O
( 3 j 2 )
Solution (a) O j 0.5236 630 6e
z ln 630 ln( 6e O
z ln( e ln 6 e
j 0.5236
j 0.5236
) ln( e
)
ln 6 j 0.5236
)
z ln 6 j 0.5236 1.7918 j 0.5236 1.867 16.29
Solution (b) 3 j 2 3.60633.69 3.606e z (3 j 2)
z e z e
e
ln 3.606 j 0.588
( 3 j 2 )
ln 3.606 j 0.588 ( 3 j 2 )
2.6718 j 4.3292
e
j 0.588
e
1.2826 j 0.588 ( 3 j 2 )
e 2.6718 j 4.3292
j 4.3292 14 . 466 e 14.166248.05
Solution (c) log( 9) log( 9) j 0.4343 log( 9) 0.9542 j1.3644 1.66555.03
(d) ln( 9) ln( 9) j ln( 9) 2.1972 j 3.1416 3.83455.03
Example Evaluate Log (1 j ) (1 j 3 ) and express the final answer in the polar form.
Solution N Log (1 j ) (1 j 3 ) N ln[(1 j ) (1 j 3 )]
N ln(1 j ) ln(1 j 3 ) N
ln(1 j 3 ) ln(1 j )
ln(260 ) 0
ln(1.414 45 ) 0
ln(2e
j1.0472
ln(1.414e
)
j 0.785
)
ln 2 j1.0472 e j1.0472 ) ln(e ) (ln 2 j1.0472) ln e N ln1.414 ln1.414 j 0.785 j 0.785 e ln(e ) ln(e ) (ln1.414 j 0.785) ln e
ln(e
N
ln 2
0.693 j1.0472 0.346 j 0.785
1.25656.51
0
0.858 66.21
0
1.464122.720
Homework 1.2 Determine the general value of the following: a. ln (3+j5) j b. log(-5) c. (6 j 4) (1 j 2)
EULER’S THEOREM • By definition
e j e j e j e j j cos j sin 2 j 2 where:
cos
e
j
e 2
j
and
sin
e
j
e j 2
j
Trigonometric Functions of Complex Numbers 1.
2.
3.
cos sin
e
j
e
j
2 e
j
e
j
j 2
e e tan j j j e e j
j
Trigonometric Functions of Complex Numbers 4.
5.
6.
e e cot j j j e e j
j
csc
sec
j 2 e
j
e
j
2 e
j
e
j
Inverse Trigonometric Functions of Complex Numbers 1.
arcsin x j ln jx 1 x
2.
arccos x j ln x
3.
arctan x j ln
1 jx 1 jx
2
x 1 2
Inverse Trigonometric Functions of Complex Numbers 4.
5.
arc cot x j ln
x j x j
1 1 x 2 arc sec x j ln x
2 1 j x 6. arc csc x j ln x
Proof of Inverse Trigonometric Functions of Complex Numbers e
sin x; sin
j
e j j 2
arcsin x e
e
j
j
j 2 e
j
e j
(e ) (e 2 2
j j
e ) j 2 xe
j
j
(e j ) 2 j 2 xe j 1 0 e
j
j 2 x 2 1 x 2
2
2 e j jx 1 x
(e ) 1 j 2 xe j
2
x
e j j 2 x
j
e j
2 j 2 x 4 4 x
2 j 2 x ( j 2 x) 4(1)(1)
2(1)
ln e j ln jx 1 x 2
1 ln jx j
1 x
j ln jx 1 x 2
arcsin x
1
2
ln jx 1 x 2
Hyperbolic Functions of Complex Numbers 1.
2.
3.
sinh x
cosh x
e e x
x
2 e e x
x
2
e tanh x e
x x
e x e x
Hyperbolic Functions of Complex Numbers 4.
5.
6.
e coth x e
x
sec hx
csc hx
x
e x e x
2
e
x
e
x
2
e
x
e
x
Inverse Hyperbolic Functions of Complex Numbers 1.
2.
3.
arcsin hy ln y
y 1
arccos hy ln y arctan hy
1 2
ln
1 y 1 y
2
; for y is
a real number ; y≥1 2
y 1
;
y 1
Inverse Hyperbolic Functions of Complex Numbers 4. arc coth y 1 ln y 1 ; y 1 2 y 1
1 1 y 2 ;0 0, -y < 0 6. arc csc hx ln y
Hyperbolic Function Identities: 1. cosh y sinh
2
y 1
2. sec h y tanh
2
y 1
2
2
3. coth y csc h y 2
4. sinh( ) 5. cosh( ) 6.
2
1
cosh sinh sinh cosh
sinh sinh cosh cosh tanh tanh tanh( ) s 1 tanh tanh
Relations Between Hyperbolic and Trigonometric Functions: 1. sinjx = jsinhx 2. cosjx = coshx 3. tanjx = jtanhx 4. sinhjx = jsinx 5. coshjx = cosx 6. tanhjx = jtanx
Example Evaluate cos(0.573 j 0.783) and express the result in polar form.
Solution cos(0.573 j 0.783) cos(0.573) cos( j 0.783) sin(0.573) sin( j 0.783) cos(0.573 j 0.783) cos(0.573) cosh(0.783) j sin(0.573) sinh(0.783) cos(0.573 j 0.783) (0.840)(1.323) j (0.542)(0.865) cos(0.573 j 0.783) 1.111 j 0.469 1.206 22.890
Example Evaluate arcsin( 3 j 4) and express the result in polar form.
Solution arcsin(3 j 4) j ln( j (3 j 4) 1 (3 j 4) 2 arcsin(3 j 4) j ln( j (3 j 4) 1 ([9 16] j[12 12]) arcsin(3 j 4) j ln( j (3 j 4) 1 (7 j 24) arcsin(3 j 4) j ln( j (3 j 4) 8 j 24 ) arcsin(3 j 4) j ln( j (3 j 4) 4.081 j 2.941)
Solution 1
1 0 2
8 j 24 (8 j 24) 2 (25.298 71.57 )
0 0 71 . 57 ( 360 ) k 8 j 24 (25.298) 2 2 k 0; 1
0 0 71 . 57 ( 0 360 ) 5.03 35.780 4.081 (25.298) 2 081 j 2.941 2 k 1; 1
0 0 71 . 57 ( 1 360 ) 2 5.03144.220 4.081 j 2.941 (25.298) 2 1
8 j 24 4.081 j 2.941
Solution () arcsin(3 j 4) j ln( j (3 j 4) 4.081 j 2.941) arcsin(3 j 4) j ln(4 j 3 4.081 j 2.941) arcsin(3 j 4) j ln(0.081 j 0.059) arcsin(3 j 4) j ln(0.136.07 0 ) arcsin(3 j 4) j ln(0.1e j 0.629 ) j ln(e ln 0.1 j 0.629 ) arcsin(3 j 4) j (ln 0.1 j 0.629) ln e arcsin(3 j 4) j (2.303 j 0.629) 0.629 j 2.303 2.38774.720
Solution ( ) arcsin(3 j 4) j ln( j (3 j 4) 4.081 j 2.941) arcsin(3 j 4) j ln(4 j 3 4.081 j 2.941) arcsin(3 j 4) j ln(8.081 j 5.941) arcsin(3 j 4) j ln(10.03143.680 ) arcsin(3 j 4) j ln(10.03e j 2.508 ) j ln(e ln10.03 j 2.508 ) arcsin(3 j 4) j (ln 10.03 j 2.508) ln e arcsin(3 j 4) j (2.306 j 2.508) 2.508 j 2.306 3.407 42.600
Example Evaluate arcsin h0.430 and express the result in polar form. O
Solution arcsin h(0.430 0 ) ln((0.4300 ) (0.430 0 ) 2 1 arcsin h(0.430 0 ) ln((0.346 j 0.2) (0.16600 ) 1 arcsin h(0.430 0 ) ln((0.346 j 0.2) (0.08 j 0.139) 1 arcsin h(0.430 0 ) ln((0.346 j 0.2) 1.08 j 0.139 arcsin h(0.430 0 ) ln((0.346 j 0.2) 1.042 j 0.067)
Solution 1
1 0 2
1.08 j 0.139 (1.08 j 0.139 ) 2 (1.089 7.334 ) 0 0 7 . 334 ( 360 ) k 1.08 j 0.139 (1.089 ) 2 2 k 0; 1
0 0 7 . 334 ( 0 360 ) 1.044 3.667 0 1.042 j 0.067 (1.089) 2 2 k 1; 1
0 0 7 . 334 ( 1 360 ) 2 1.044 183 .667 0 1.042 j 0.067 (1.089) 2 1
1.08 j 0.139 1.042 j 0.067
Solution ( ) arcsin h(0.4300 ) ln((0.346 j 0.2) 1.042 j 0.067) arcsin h(0.4300 ) ln(1.388 j 0.267) arcsin h(0.430 ) ln(1.41310.89 ) 0
0
arcsin h(0.4300 ) ln(1.413e j 0.19 ) ln(e ln1.413 j 0.19 ) arcsin h(0.4300 ) (ln 1.413 j 0.19) ln e arcsin h(0.430 ) (0.346 j 0.19) 0.39528.77 0
0
Solution ( ) arcsin h(0.430 0 ) ln((0.346 j 0.2) 1.042 j 0.067) arcsin h(0.430 ) ln(0.696 j 0.133) 0
arcsin h(0.430 ) ln(0.709169.18 ) 0
0
arcsin h(0.430 0 ) ln(0.709e j 2.953 ) ln(e ln 0.709 j 2.953 ) arcsin h(0.430 0 ) (ln 0.709 j 2.953) ln e arcsin h(0.430 ) (0.344 j 2.953) 2.97396.64 0
0
Example Evaluate sinh 0.346 j 0.548 and express the result in polar form.
Solution sinh 0.346 j 0.548 sinh(0.346) cosh( j 0.548) cosh(0.346) sinh( j 0.548) sinh 0.346 j 0.548 sinh(0.346) cos(0.548) j cosh(0.346) sin(0.548) sinh 0.346 j 0.548 (0.353)(0.854) j (1.0604)(0.521) sinh 0.346 j 0.548 0.301 j 0.552 0.629 61.400
Seatwork 1.3 Evaluate the following and express the result in polar form. 1. cos0.492 j 0.942
2. a rc c ot j 6 3. sinh 0.5 j 0.75
Cauchy-Riemann Equations • It can be obtain from the derivative of any of the following formulas:
u v v u i i dz x x y y
dw and
v v u u i i dz y x x y
dw
Cauchy-Riemann Equations Example Show that sin(z) is an entire function.
Cauchy-Riemann Equations Solution
sin( z ) sin( x jy ) sin( x) cos( jy ) cos( x) sin( jy )
sin( x) cosh( y ) j cos( x) sinh( y ) u sin( x) cosh( y ) v cos( x) sinh( y ) u cos( x) cosh( y ) x u sin( x) sinh( y ) y v sin( x) sinh( y ) x v cos( x) cosh( y ) y
Cauchy-Riemann Equations Example Consider the function w = 1/z
Cauchy-Riemann Equations w u jv
Solution w u jv u ( x, y )
1 x jy x x y 2
x x y 2
2
iy x y 2
, v( x, y ) 2
2
y x y 2
2
2 2 u x 2 y 2 2 x 2 y x 2 2 2 2 x x y x y 2 2 2 2 2 2 2 v u x y 2 y y x 2 2 2 2 2 2 y x y x y x 0 2 xy 2 xy v 2 x x y 2 2 x 2 y 2 2
0 2 xy 2 xy u v 2 2 2 2 2 2
Cauchy-Riemann Equations Example Find the derivative of the following using Cauchy-Reimann equations: a. d
sin( z)
dz b. d 1 dz z
Cauchy-Riemann Equations Solution: a.
u v sin( z ) j x x dz d d dz d dz
sin( z ) cos( x) cosh( y) j sin( x) sinh( y ) sin( z ) cos( x jy) cos( z )
Cauchy-Riemann Equations Solution: b.
d 1
y x 2
2
2 jxy
2 2 2 2 2 2 dz z x y x y d 1
1
1
2 2 dz z z x jy
