Complex Numbers

What is Complex Number?

Complex Number is a combination (sum or
difference) of real and imaginary numbers.

Definition of Complex Numbers

A complex number, z, is a number in the
form of a + bi or x +jy, where i or j = -1.
a or x is called the real part of z, and b or y
is called the imaginary part of z.
The standard form of a complex number
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

= 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

where: r = magnitude or amplitude

= 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 45 6.708
2

3
tan 26.56
6
z r 6.708 26.56
1

Solution
a.
exponential form

r 6 3 45 6.708
2

3
tan 26.56
0.4636
180
6
j
j 0.4636
z re 6.708e
1

Solution
a.
trigonometric form

r 6 3 45 6.708
2

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

0.524

180
j 0.524
6e

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

4 n 1

i,

4n2

1,

4 n 3

i ,
i i i i, i i i i, i
8
4
4
4n4
4
2
2
1,
i i i 1, i i i 1, i
3

Example
Find the equivalent of the following:
a.
b.
c.
d.

i1995
i2006
i1988
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 =
y 2.
3. If (x1 + iy1)(x2 + iy2) = 0, then at least one
of the factors is zero, that is , x1 + 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 x1 iy1
x1 iy1 x2 iy2

z2 x2 iy2 x2 iy2 x2 iy2

z1 x1 x2 y1 y2
x2 y1 x1 y2

i 2
2
2
z2
x2 y2
x2 y22

Arithmetic Operations in
Rectangular Form
e. Extraction of Square Roots:

x jy a jb
or

x jy

1
2

1
2

r ( 360 k ) 2
0

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 r1 cos1 j sin 1 r11 ,
z2 r2 cos 2 j sin 2 r2 2 .
z1 z2 r11 r2 2

z1 z2 r1r2 cos1 j sin 1 cos 2 j sin 2

z1 z2 r1r2 [(cos1 cos 2 sin 1 sin 2 )
j (sin 1 cos 2 cos1 sin 2 )]
z1 z2 r1r2 [cos(1 2 ) j sin(1 2 )]
z1 z2 r1r21 2

Multiplication of two Exponential

Forms of Complex Numbers
z1 r1e

j 1

z2 r2e

j 2

z1 z2 r1e

j 1

z1 z2 r1r2e

r e

j 2

j (1 ( 2 ))

Division of two Polar Forms of

Complex Numbers
z1 r11 r1 cos1 j sin 1 cos 2 j sin 2

z2 r2 2 r2 cos 2 j sin 2 cos 2 j sin 2

z1 r1 cos1 cos 2 sin 1 sin 2
sin 1 cos 2 cos1 sin 2

j

2
2
z2 r2
cos 2 sin 2
cos2 2 sin 2 2

z1 r1
cos(1 2 ) j sin(1 2 )
z2 r2
z1 r1
1 2
z2 r2

Division of two Exponential Forms

of Complex Numbers
z1 r1e

j 1

z2 r2e

j 2
j 1

z1 r1e
r1 j (1 2 )

e
j 2
z2 r2e
r2

Example
Perform the indicated operations:
O
j 0.752
O
5 25
a. 6 j 7 10cjs 30 10e

b. 5 j 3 630 O

5 j4
c.
3 j4

Solution
a.

6 j 7 10cjs 30 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

Solution
b. rectangular form

5 j35.196 j3
2
(5)(5.196) j (3)(3) ( j3)(5.196) (5)( j3)
25.98 9 j 15.588 15
16.98 j30.588 34.9860.96

Solution
b. polar form

5.83130.96 630
(5.831)(6)30.96 30
34.986 60.96

Solution
c.

5 j 4 3 j 4 15 16 j12 j 20

3 j 4 3 j 4 9 16 j12 j12
31 j8
1.24 j 0.32
25

Example
Evaluate the square root of (3+j4).

Solution
x jy 3 j 4

x jy

3 j4

x j 2 xy j y 3 j 4
2

x y j 2 xy 3 j 4
2

real : x y 3
2

eqn.1
4 2
imaginary : 2 xy 4; y

2x x

Solution
2

2
x 3
x
2 4
2
x x 2 3 x
2

x
x

2 2

4 3x

2 2

3x 4 0
2

Solution
Using quadratic formula:

3 9 16 3 25 3 5

x
2
2
2
()
35 8
2
4
x
2
2
x 4 2
2

Solution
Using quadratic formula:

( )
35 2
x

1
2
2
x 1 imaginary(drop)
2
2
x 2 and y
1
x 2
2

Solution
2 j 3 j4
first root : 2 j
sec ond root : 2 j

Solution (alternative)
360 k
x jy r

3 j 4 553.13
first root : k 0
1
2

1
2

53.13 360 (0)

5

2.236 26.56 2 j
1
2

Solution (alternative)
sec ond

root : k 1

53.13 360 (1)

5

2.236 206 .56

2 j
1
2

Seatwork 1.1
1. Find the sum and difference of
O
j 0.752
5cjs30 2e
3 j5
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 Moivres Theorem
j
Let z re r
2
2 j 2
2
r 2 ;
successive powers, z r e
3
3 j 3
3
z

r
e

,...
In general,
n
n jn
n
z r e r n ,
letting r = 1,

cos j sin

cos n j sin n

or in the abbreviated form.

Example
2
2

1. Simplify cos 2 j sin 2 cos 6 j sin 6

3
cos3 j sin 3 cjs10

3
2. Evaluate 1
3

j
2
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

10
9
cos j sin cos j sin
16

cos j sin

cos j sin 19

1
1

3
cos j sin cos3 j sin 3

Solution
2. 1

3
2 j 2

1 j 3 cos j sin
2

cos 1 ; cos1 1 60
2
2
sin 3 ; sin 1 3 60
2
2
cos 60 j sin 60 3 cos180 j sin 180
1

Roots of Complex Numbers

Let

z re

j ( 2 k )

r k 360

Then
n

1
n

1
n

z z r e
1
n

2 k
j

k 360
n

, k 0,1,2,...

Roots of Complex Numbers

0
360 0 k
360 k
j sin

r cos
n
n

0
0

360 k
360 k
n
j sin

Wk r cos
n
n

1
n

where:
k = 0,1,2,(n-1)
W0 = is the principal value or root, and must be
a positive angle.

Example
1. Find the three roots of 125
2. Find the four roots of 16 20

Solution
1.

125 1250
first root :
k 0
0
0

0 360 (0)
0 360 (0)
3
j sin

W0 125 cos
3
3

W0 5cos 0 j sin 0 5cjs 0 50

Solution
1.

sec ond
k 1

root :

0
0

0 360 (1)
0 360 (1)
3
j sin

W1 125 cos
3
3

W1 5cos120 j sin 120 5cjs120 5120

Solution
1.

third root :
k 2

0
0

360
(
2
)
0

360
(2)
3
j sin

W2 125 cos
3
3

W2 5cos 240 j sin 240 5cjs 240 5240

Solution
2.

16 20
first root :
k 0
0

20 360 (0)
20 360 (0)
W0 16 cos

j sin
4
4

W0 2cos(5) j sin( 5) 2cjs (5) 2 5 2355
4

Solution
2.

sec ond root :

k 1
20 360(1)
20 360(1)
W1 16 cos
j sin

4
4

W1 2cos85 j sin 85 2cjs85 285
4

Solution
2.

third root :
k 2
20 360(2)
20 360(2)
W2 16 cos
j sin

4
4

W2 2cos175 j sin 175 2cjs175 2175
4

Solution
2.

fourth root :
k 3
20 360(3)
20 360(3)
W3 16 cos
j sin

4
4

W3 2cos 265 j sin 265 2cjs 265 2265
4

Homework 1.1
1. Find the three roots of 1230

2. Simplify 2 cos

j sin

Exponential and Trigonometric

Functions of Complex Numbers
From Eulers 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
e

j ( z1 z 2 )

j ( z1 z 2 )

e
j2
e
2

j ( z1 z 2 )

j ( z1 z 2 )

Eqn. 1

Exponential and Trigonometric

Functions of Complex Numbers
e
e
e

j ( z1 z 2 )

j ( z1 z 2 )

cos z1 j sin z1 cos z2 j sin z2

j ( z1 z 2 )

jz1

jz 2

cos z1 cos z2 sin z1 sin z2

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 )

j ( z1 z 2 )

cos z1 j sin z1 cos z2 j sin z2

j ( z1 z 2 )

jz1

jz 2

cos z1 cos z2 sin z1 sin z2

j sin z1 cos z2 cos z1 sin z2

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
Let z1 x, z 2 jy

Then z1 z 2 x jy

Exponential and Trigonometric

Functions of Complex Numbers
sin( x jy ) sin x cos jy cos x sin jy
y

e e
e e
sin( x jy ) sin x
cos x
2
j2
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
y

Hyperbolic Functions of Complex

Numbers
hyperbolic functions for the complex number z

e e
sinh z
2
z

e e
, cosh z
2
z

Hyperbolic Functions of Complex

Numbers
cosh z sinh z 1
sinh( z1 z2 ) sinh z1 cosh z2 cosh z1 sinh z2
2

cosh(z1 z2 ) cosh z1 cosh z2 sinh z1 sinh z2

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)
b. cosh( j 0.78)
c.

sinh( 0.942 j 0.429)

d.

cos(0.942 j 0.429)

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

j
0
.
429
)

cos
(0.942 j 0.429) 1
c.

Logarithms of Complex Numbers

Express the complex number z = x + jy in the
general exponential form

z re

j ( 2 k )

where 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 of 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

Log ( A ) Log A e

j / 180

j / 180

Log ( A ) LogA Loge

Log ( A ) LogA j / 180 Loge
Log ( A ) LogA j / 180 0.4343

Logarithm and Natural Logarithm of

a Complex Numbers

Ln( A ) Ln A e

j / 180

j / 180

Ln( A ) LnA Lne

Ln( A ) LnA j / 180 Lne
Ln( A ) LnA j / 180 1

Logarithm and Natural Logarithm of

a Complex Numbers

Log ( N ) Log N180

Log ( N ) LogN j180 / 180 Loge
Log ( N ) LogN j 0.4343

Logarithm and Natural Logarithm of

a Complex Numbers

Ln( N ) LnN180
Ln( N ) LnN j180 / 180 Lne
Ln( N ) LnN j 1

Example
Determine the general value of the
following :
O
a. ln 630
(3 j 2)
b.(3 j 2)
c. Log(-9)
d. Ln(-9)

Solution
(a)
630 O 6e j 0.5236
z ln 630 ln( 6e
O

j 0.5236

z ln( e ln 6 e 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 j 0.588 e ln 3.606 j 0.588
z (3 j 2) (3 j 2)

z e

ln 3.606 j 0.588 ( 3 j 2 )

1.2826 j 0.588 ( 3 j 2 )

e 2.6718 j 4.3292

z e 2.6718e j 4.3292 14.466e j 4.3292 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 3.834 55.03

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 )
ln[(1 j ) N (1 j 3 )]
N ln(1 j ) ln(1 j 3 )
ln(1 j 3 )
ln(2600 )
ln(2e j1.0472 )
N

0
ln(1 j )
ln(1.414 45 ) ln(1.414e j 0.785 )
ln(e ln 2 e j1.0472 )
ln(e ln 2 j1.0472 )
(ln 2 j1.0472) ln e
N

ln1.414
j 0.785
ln1.414 j 0.785
ln(e
e
) ln(e
) (ln 1.414 j 0.785) ln e
0.693 j1.0472
1.25656.510
0
N

1
.
464

122
.
72
0.346 j 0.785 0.858 66.210

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)

EULERS THEOREM
By definition

e j e j
cos j sin
2

where:

e j e j
cos
2

and

e j e j
j
j2

sin

e
j2

Trigonometric Functions of
Complex Numbers
1.

2.

3.

cos

sin

e
2

e
j2
j

e e
tan j j
j
e e

Trigonometric Functions of
Complex Numbers
4.

5.

6.

e e
cot j j
j
e

csc

sec

j2
j
e

2
j
e

Inverse Trigonometric Functions of

Complex Numbers

1.

arcsin x j ln jx 1 x

2.

arccos x j ln x x 1

jx
3. arctan x j ln
1 jx

Inverse Trigonometric Functions of

Complex Numbers
4.

5.

arc cot x j ln

x j
x j

1 1 x2
arc sec x j ln

1
6.

arc csc x j ln

Proof of Inverse Trigonometric

Functions of Complex Numbers
e j e j
sin x; sin
j2
arcsin x
e

e
j2

(e ) (e

e ) j 2 xe

(e ) 1 j 2 xe

(e j ) 2 j 2 xe j 1 0
e j

e j

j2x 2 1 x 2

e j jx 1 x 2

e j e j j 2 x
j

e j

j2x 4 4x 2

j 2 x ( j 2 x) 2 4(1)(1)

2(1)

ln e j ln jx 1 x 2

j ln jx 1 x 2
1
ln jx 1 x 2
j
1
arcsin x ln jx 1 x 2
j

Hyperbolic Functions of Complex

Numbers
1.

e
sinh x

2.

e
cosh x

3.

e
tanh x
e

x
x

e
2

e
2

e
x
e

Hyperbolic Functions of Complex

Numbers
4.

e
coth x
e

x
x

e
x
e

5.

2
sec hx x
x
e e

6.

2
csc hx x
x
e e

Inverse Hyperbolic Functions of

Complex Numbers
1.

arcsin hy ln y

y 1

2.

arccos hy ln y

3.

1 1 y
arctan hy ln
2 1 y

; for y is

a real number
;
y1
2

y 1
;

y 1

Inverse Hyperbolic Functions of

Complex Numbers
4. arc coth y 1 ln y 1 ; y 1

y 1

1 1 y2
5. arc sec hy ln

;0<y1

1 1 y2
6. arc csc hx ln

; +y > 0, -y < 0

Hyperbolic Function Identities:

1.

cosh2 y sinh 2 y 1

2.

sec h y tanh y 1

3.

coth y csc h y 1

4.

sinh( ) sinh cosh cosh sinh

cosh( ) cosh cosh sinh sinh

tanh

tanh

6. tanh( ) s
1 tanh tanh
5.

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
result in polar form.

j 4) and express the

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
2

1
0 2

8 j 24 (8 j 24) (25.298 71.57 )

71.57 0 (k 360 0 )

8 j 24 (25.298)
2

k 0;
1
2

71.57 0 (0 360 0 )
5.03 35.780 4.081 j 2.941
(25.298)
2

k 1;
1
2

71.57 0 (1 360 0 )
5.03144 .22 0 4.081 j 2.941
(25.298)
2

1
2

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(0.4300 ) ln((0.4300 ) (0.4300 ) 2 1
arcsin h(0.4300 ) ln((0.346 j 0.2) (0.16600 ) 1
arcsin h(0.4300 ) ln((0.346 j 0.2) (0.08 j 0.139) 1
arcsin h(0.4300 ) ln((0.346 j 0.2) 1.08 j 0.139
arcsin h(0.4300 ) ln((0.346 j 0.2) 1.042 j 0.067)

Solution
1
2

1
0 2

1.08 j 0.139 (1.08 j 0.139 ) (1.089 7.334 )

7.334 0 (k 360 0 )

1.08 j 0.139 (1.089 )

2

k 0;
1
2

7.334 0 (0 360 0 )
1.044 3.667 0 1.042 j 0.067
(1.089 )
2

k 1;
1
2

7.334 0 (1 360 0 )
1.044 183 .667 0 1.042 j 0.067
(1.089 )
2

1
2

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(0.4300 ) ln(1.41310.890 )
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

Solution
( )
arcsin h(0.430 ) ln((0.346 j 0.2) 1.042 j 0.067)
0

arcsin h(0.430 ) ln(0.696 j 0.133)

0

arcsin h(0.4300 ) ln(0.709169.180 )

arcsin h(0.4300 ) ln(0.709e j 2.953 ) ln(e ln 0.709 j 2.953 )
arcsin h(0.4300 ) (ln 0.709 j 2.953) ln e
arcsin h(0.430 ) (0.344 j 2.953) 2.97396.64
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. arc cot 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:

dw u
v v u

i
i
dz x
x y
y
and

dw v
v u
u

i
i
dz y
x x
y

Cauchy-Riemann Equations
Example
Show that sin(z) is an entire function.

Cauchy-Riemann Equations
sin( z ) sin( x jy ) sin( x) cos( jy ) cos(x) sin( jy )
Solution
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
1
x jy
x
iy
w u jv 2

x y 2 x2 y 2
x
y
u ( x, y ) 2
,
v
(
x
,
y
)

x y2
x2 y 2
w u jv

Solution

u
x2 y2 2x2
y2 x2

2
2
2
2
2
x
x y
x y2

v
x2 y 2 2 y 2
y 2 x2
u

2
2
y
x
x2 y 2
x2 y 2
v
0 2 xy
2 xy
2

2
2
x
x y2
x2 y2

u
0 2 xy
2 xy
2

2
y
x y2
x2 y 2

v
x

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.

d
u
v
sin( z ) j
dz
x
x
d
sin( z ) cos(x) cosh(y) j sin( x) sinh( y)
dz
d
sin( z ) cos(x jy ) cos(z )
dz

Cauchy-Riemann Equations
Solution:
b.

d 1
y x
2
dz z x y 2
2

x
2

2 jxy
2

d 1
1
1
2

2
dz z
z
x jy

2 2

Seatwork 1.4
1. Show that following are an entire function.

f ( z) e

(a)

f ( z ) cosh(z )
(b)

2. Find the derivative of the following using

Cauchy-Reimann equations:

f ( z) e z

f ( z ) cosh(z )

(a)

(b)