Complex Exponential function as
z
e
We also need to give a meaning to the expression
z
e
when
z
i
x
y
is a complex
number. Using the Taylor series for
z
e
as our guide, we define
0
2
3
4
1
!
!
!
2
3
4
!
z
n
n
e
z
z
z
z
z
n
(7)
and it turns out that this complex exponential function has the same properties as the real exponential
function. In particular, it is true that
1
2
1
2
z
z
z
z
e
e
e
(8)
If we put
z
y
i
, where
y
is a real number, in Equation (7), and use the facts that
2
3
2
4
2
2
5
3
2
,
,
,
,
1
1
i
i
i
i
i i
i
i
i
i
i
i
we get
2
3
4
5
2
3
4
5
2
4
3
5
!
!
!
!
!
!
!
!
cos
sin .
!
!
1
2
3
4
5
!
!
1
2
3
4
5
1
2
4
3
5
z
i
i
i
i
i
i
i
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
y
i
i
i
y
e
(7.1)
Here we have used the Taylor series for
cos
y
and
sin
y
. The result is a famous formula
called
Euler’s formula
:
cos
sin .
i
y
y
i
y
e
(9)
Combining
Euler’s
formula (9) with (8), we get
cos
sin
x
y
x
y
x
i
i
e
e
i
e
e
y
y
(10)
Thus,
z
e
represents complex function with real part
cos
x
e
y
and the imaginary part
sin
x
e
y
. The exponential function with arbitrary (real) base is defined via the exponential function
as
IIUM, Faculty of Engineering,
Department Engineering in Science
Engineering Mathematics I
Semester 1, 2021/2022
Chapter I:
Complex Numbers
Lecturer
Associate Professor Dr. Abdurahim Okhunov
23
ln
z
z
a
a
e
.
The power function can also be defined via the exponential function. In this course we will
work just with integer powers of
z
: So, let
n
be an integer.
i
n
n
n
n
i
r e
=
z
=
r
e
Example 15.
Evaluate:
a)
i
e
and
b)
1
2
i
e
Solution:
a)
From (8) we have
cos
sin
1
0
1
i
e
i
i
.
b)
Using (9) we get
1
1
1
1
2
2
c
0
1
2
o
2
s
sin
i
i
e
i
i
i
e
e
e
e
e
.
Using the trigonometric representation, the formulas for multiplication and division of two
complex numbers become easier than when the Cartesian form of complex numbers is used. If
1
z
is a
complex number with modulus
1
r
and phase
1
and
2
z
is a number with modulus
2
r
and phase
2
,
then
1
2
1
1
2
2
1
2
1
2
1
2
1
2
1
2
1
2
+
i
r
r
r
cos
+ sin
cos
+ sin
cos
+
+ sin
z
=
i
i
z
i
r
+
r r e
with the agreement that if
1
2
is larger than
2
; then
2
is subtracted from this sum. This gives
us an
easy formula for the
th
n
power
of a complex number
=
cos
r
n
z
+
i
si
which we will
discuss below.
Finally, we note that
Euler’s formula
(8) provides us with an easier method of proving De
Moivre’s Theorem:
cos
sin
cos
sin
n
i
n
n
r
re
r
n
y
n
i
y
i
(10.a)
IIUM, Faculty of Engineering,
Department Engineering in Science
Engineering Mathematics I
Semester 1, 2021/2022
Chapter I:
Complex Numbers
Lecturer
Associate Professor Dr. Abdurahim Okhunov
24
De Moiver’s Theorem:
If
x
y
r e
i
z
i
and
n
is a natural number, then
n
n
n
n
n
r e
z
r
i
e
x
y
i
i
(11)
Example 16
(The natural number power of a Complex number)
:
Use De Moivre’s theorem to find
10
1
i
. Write the answer in exact rectangular form.
Solution:
10
10
10
10
0
0
cos
s
1
.
0
in
0
0
0
45
45
45
i
i
0
i
450
450
Rectangular f
e
o
2
2
32
32
32
3
i
i
i
i
2
m
e
r
e
=
=
=
=
=
Matched Problem:
1.
Use De Moivre’s theorem to find
5
1
3
i
. Write the answer in exact rectangular form.
2.
Use De Moivre’s theorem to find
4
1
3
i
. Write the answer in exact rectangular form.
The Powers and
n
th
Root of complex numbers
The
th
n
Root of complex numbers are:
n
n
n
=
cos + sin
n
n
=
cos
+ sin
i
z
i
r
r
For example
2
2
2
2
3
3
3
3
4
4
4
4
1.
;
2.
;
3.
.
i
i
i
i
i
i
i
i
r e
r e
r e
r e
r e
r
x
e
y
x
y
x
i
y
Explore – Discuss
.
If
i
z
r e
, then use
De Moivre’s theorem
to show that
1 2
2
i
r e
is a square root of
z
and
1 3
3
i
r e
is a cube root of
z
. We can proceed in the same way as in
Explore – Discuss
to show that
1
i
n
n
r e
is an
th
n
root of
i
r e
,
n
a natural number:
IIUM, Faculty of Engineering,
Department Engineering in Science
Engineering Mathematics I
Semester 1, 2021/2022
Chapter I:
Complex Numbers
Lecturer
Associate Professor Dr. Abdurahim Okhunov
25
1
1
i
i
i
n
n
n
n
n
n
n
r e
r
e
r e
But we can do even better than this.
Theorem 2
(
th
n
– Root Theorem
)
:
For
n
a positive integer greater, then
1
1
1
cos
sin
n
n
n
n
r
r
e
z
i
i
(12)
But this formula gives us only first root, therefore
Theorem 3
(
th
n
– Root Theorem
)
:
For
n
a positive integer greater, then
1
1
1
is number o
, , ,
,
f roo
2
~
2
ts
.
0 1
1
cos
si
2
2
,
,
n
n
n
n
k
n
n
k
i
n
n
k
k
r
r
n
z
i
e
n
n
n
(13)
Are the
n
distinct
th
n
roots of
i
r e
, and there are no others. For example
1
3
3
3
2
8
8
,
8
.
Example 17
(Roots of Complex number)
:
Find all the values of
x
for
3
8
x
, and locate these values in a complex plane.
Solution:
First, we have to present this complex form as
z
i
a
b
. So we have that
8
0
z
i
, this we can
write in polar form:
2
8
0
64
8
z
r
and
180
,
where
3
n
, and
,
0
2
,
1
k
, then
0
180
Im
Re
Im
8
0
180
IIUM, Faculty of Engineering,
Department Engineering in Science
Engineering Mathematics I
Semester 1, 2021/2022
Chapter I:
Complex Numbers
Lecturer
Associate Professor Dr. Abdurahim Okhunov
26
1
1
1
3
3
3
1
1
1
3
3
3
cos
sin
cos
sin
;
c
2
2
0
3
3
3
3
3
3
2
2
2
2
1
3
3
,
,
os
sin
cos
sin
;
0
1
3
3
3
1
3
0
z
z
k
r
r
k
i
r
r
i
i
k
i
1
1
1
3
3
3
cos
sin
cos
sin
.
2
2
4
4
1
3
3
3
3
3
,
3
2
2
i
r
z
i
r
Example 18
(Finding all Sixth Roots of a Complex number)
:
Find six distinct sixth roots of
1
3
i
, and plot them in a complex plane.
Solution:
The first we will write
1
3
i
in polar form:
2
2
ta
0
n
12
1
3
1 3
4
2
3
1
z
r
3
, then
60
1
3
2
i
z
i
e
.
Using the
th
n
– root theorem, all six roots are given by
0
0
0
0
120
360
6
6
1 6
20
60
1 6
2
,
, , , , ,
2
0 1 2 3 4 5
k
k
i
i
e
k
e
Thus,
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
20
60
20
20
60
80
20
60
140
20
60
200
20
60
26
0
1 6
1 6
1
1 6
1 6
2
1 6
1 6
3
1 6
1 6
4
1 6
1 6
5
1 6
1 6
0
20
60
320
;
;
.
;
;
;
1
2
3
4
i
i
i
i
i
i
i
i
i
i
i
6
i
5
2
2
2
2
2
2
2
2
2
2
2
e
e
e
e
e
e
e
e
e
e
e
w
w
w
w
w
w
e
2
y
x
w
2
w
1
w
3
w
4
w
5
w
6
Radius
2
1
/
6
IIUM, Faculty of Engineering,
Department Engineering in Science
Engineering Mathematics I
Semester 1, 2021/2022
Chapter I:
Complex Numbers
Lecturer
Associate Professor Dr. Abdurahim Okhunov
27
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