The document covers the theory and application of complex numbers, including their properties, polar forms, and methods for calculating powers and roots using De Moivre's theorem. It provides examples and exercises to illustrate how to find complex roots, applications of complex numbers in quadratic equations, and transformations between rectangular and polar forms. Key mathematical concepts such as the quadratic formula, various powers of complex numbers, and their geometric interpretations are also discussed.

1. Complex
Variables &
Transforms
Instructor: Dr. Naila Amir
MATH- 232

2. Complex numbers
▪ Complex Numbers and Their Properties
▪ Complex Plane
▪ Polar Form of Complex Numbers
▪ Powers of Complex Numbers
▪ Roots of Complex Numbers
▪ Sets of Points in the Complex Plane
▪ Applications

3. We can find integer powers of a complex number 𝑧 by using results of multiplication
and division in polar form. If 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃), then
𝑎 𝑧2
= 𝑧𝑧 = 𝑟2
cos(𝜃 + 𝜃) + 𝑖 sin (𝜃 + 𝜃) = 𝑟2
cos 2𝜃 + 𝑖 sin 2𝜃 ,
𝑏 𝑧3
= 𝑧2
𝑧 = 𝑟3
cos(2𝜃 + 𝜃) + 𝑖 sin (2𝜃 + 𝜃) = 𝑟3
cos 3𝜃 + 𝑖 sin 3𝜃 ,
𝑐 𝑧4
= 𝑧3
𝑧 = 𝑟4
cos(3𝜃 + 𝜃) + 𝑖 sin (3𝜃 + 𝜃) = 𝑟4
cos 4𝜃 + 𝑖 sin 4𝜃 ,
and so on. Also,
1
𝑧
=
1
𝑟
cos( 0 − 𝜃) + 𝑖 sin( 0 − 𝜃) = 𝑟−1
cos( − 𝜃) + 𝑖 sin( − 𝜃) .
Similarly,
1
𝑧2
= 𝑧−2 = 𝑟−2 cos( − 2𝜃) + 𝑖 sin( − 2𝜃) .
Integer Powers of 𝒛

4. Continuing in the same manner, we obtain a formula for the 𝑛th power of 𝑧 for any
integer 𝑛 as:
𝑧𝑛
= 𝑟𝑛
cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃 = 𝑟𝑛
cis 𝑛𝜃 = 𝑟𝑛
𝑒𝑖𝑛𝜃
. ∗
Eq. (∗) is the general form of the De Moivre's theorem which is stated as:
“If 𝑧 = cos 𝜃 + 𝑖 sin 𝜃 such that 𝑧 = 𝑟 = 1 the for any integer 𝑛
𝑧𝑛
= cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃 = cis 𝑛𝜃 = 𝑒𝑖𝑛𝜃
. ”
Note: This theorem is also valid if 𝑛 is a rational number.
Integer Powers of 𝒛

5. Determine 𝑧3 if 𝑧 =
3
2
+ 𝑖
1
2
.
Solution:
For the present case we have:
𝑧 = 1 and Arg 𝑧 = 𝜃 =
𝜋
6
.
Thus,
𝑧3
=
3
2
+ 𝑖
1
2
3
= cos
𝜋
6
+ 𝑖 sin
𝜋
6
3
= cos
3𝜋
6
+ 𝑖 sin
3𝜋
6
= cos
𝜋
2
+ 𝑖 sin
𝜋
2
= 𝑖.
Example

6. De Moivre’s Theorem
▪ De Moivre's theorem, named after the French mathematician Abraham de Moivre
is used the find the powers and roots of complex numbers.
▪ De Moivre's Theorem states that the power of a complex number in polar form is
equal to raising the modulus to the same power and multiplying the argument
by the same power.
▪ De Moivre's theorem can be derived from Euler's equation, and is important
because it connects trigonometry to complex numbers.
▪ In general, for any complex number 𝑧 and any integer 𝑛, the following is true:
If 𝑧 = 𝑟 cos 𝜃 + 𝑖 sin 𝜃 then 𝑧𝑛
= 𝑟𝑛
cos 𝑛 𝜃 + 𝑖 sin 𝑛 𝜃 .

7. Given 𝑧 = 1 + 𝑖 3, find
𝑖 𝑧2 𝑖𝑖 𝑧5 𝑖𝑖𝑖 𝑧7.
Solution:
For the present case note that:
𝑧 = 1 + 3 = 2,
and
tan 𝜃 =
3
1
⇒ 𝜃 =
𝜋
3
.
Thus,
𝑧 = 2 cos
𝜋
3
+ 𝑖 sin
𝜋
3
.
Example

8. Since,
𝑧 = 2 cos
𝜋
3
+ 𝑖 sin
𝜋
3
,
so
𝑖 𝑧2
= 4 cos
2𝜋
3
+ 𝑖 sin
2𝜋
3
= 4
−1
2
+ 𝑖
3
2
= −2 + 2 3𝑖.
𝑖𝑖 𝑧5
= 25
cos
5𝜋
3
+ 𝑖 sin
5𝜋
3
= 32 cos
−𝜋
3
+ 𝑖 sin
−𝜋
3
= 16 − 16 3𝑖.
𝑖𝑖𝑖 𝑧7
= 27
cos
7𝜋
3
+ 𝑖 sin
7𝜋
3
= 128 cos
𝜋
3
+ 𝑖 sin
𝜋
3
= 64 + 63 3𝑖.
Example

9. Book: A First Course in Complex Analysis with Applications by Dennis G.
Zill and Patrick D. Shanahan.
Chapter: 1
Exercise: 1.3
Q # 1 – 38.
Practice Questions

10. Applications

11. Complex Numbers
Book: A First Course in Complex Analysis with Applications by
Dennis G. Zill and Patrick D. Shanahan.
• Chapter: 1
• Sections: 1.4, 1.6

12. Applications of De Moivre’s Theorem
▪ To express cos 𝑛𝜃 and sin 𝑛𝜃 as finite sums of trigonometric
functions of 𝜃, where 𝑛 is a positive integer.
▪ To express powers of cos 𝜃 (or sin 𝜃) in a series of cosines (or
sines) of multiples 𝜃.
▪ To find 𝑛th roots of a complex number.

13. Roots of a Complex Number
Suppose 𝑟 cos 𝜃 + 𝑖 sin 𝜃 and 𝜌 cos 𝜑 + 𝑖 sin 𝜑 are polar forms of the complex
numbers 𝑧 and 𝜔 respectively. Then, the equation: 𝜔𝑛 = 𝑧, becomes:
𝜌𝑛
cos 𝑛𝜑 + 𝑖 sin 𝑛𝜑 = 𝑟 cos 𝜃 + 𝑖 sin 𝜃 (1)
From (1), we can conclude that:
𝜌𝑛 = 𝑟 (2)
and
cos 𝑛𝜑 + 𝑖 sin 𝑛𝜑 = cos 𝜃 + 𝑖 sin 𝜃 (3)
From (2), we define:𝜌 = 𝑟1/𝑛 to be the unique positive 𝑛th root of the positive
real number 𝑟. From (3), the definition of equality of two complex numbers
implies that:
cos 𝑛𝜑 = cos 𝜃 and sin 𝑛𝜑 = sin 𝜃 .
These equalities, in turn, indicate that the arguments 𝜃 and 𝜑 are related bythe
equation 𝑛𝜑 = 𝜃 + 2𝑘𝜋, where 𝑘 is an integer.

14. Roots of a Complex Number
Thus,
𝜑 =
𝜃 + 2𝑘𝜋
𝑛
.
As 𝑘 takes on the successive integer values 𝑘 = 0, 1, 2, … , 𝑛 − 1, we obtain 𝑛
distinct 𝑛th roots of 𝑧; these roots have the same modulus 𝑟1/𝑛
but different
arguments. Thus, the 𝑛 𝑛th roots of a non-zero complex number 𝑧 are given by:
𝜔𝑘 = 𝑧1/𝑛 = 𝑟1/𝑛 cos
𝜃 + 2𝑘𝜋
𝑛
+ 𝑖 sin
𝜃 + 2𝑘𝜋
𝑛
, (4)
where 𝑘 = 0, 1, 2, . . . , 𝑛 − 1. These 𝑛 values lie on a circle of radius 𝑟1/𝑛 with
center at the origin and constitute the vertices of a regular polygon of 𝑛 sides. The
value of 𝑧1/𝑛
obtained by taking the principal value of arg 𝑧 and 𝑘 = 0 in (4) is
called the principal 𝒏th root of 𝜔 = 𝑧1/𝑛.

15. Example
Determine the four fourth roots of 𝑧 = 2(1 + 𝑖).

16. Example
Note that, since Arg(𝑧) = 𝜋/4, so we have:
𝜔0 ≈ 1.1664 + 𝑖 0.2320,
is the principal fourth root of 𝑧 = 2(1 + 𝑖).
As shown in figure, the four roots lie on a circle
centered at the origin of radius 𝑟 =
4
2 ≈ 1.19
and are spaced at equal angular intervals of
2𝜋/4 = 𝜋/2 radians, beginning with the root
whose argument is 𝜋/16.

17. Example
Determine the cube roots of 𝑧 = 𝑖.

18. Example
Note that, since Arg(𝑧) = 𝜋/2, so we have:
𝜔0 ≈ 0.8660 + 0.5𝑖,
is the principal cubic root of 𝑧 = 𝑖.
As shown in figure, the three roots lie on a circle
centered at the origin of radius 𝑟 ≈ 1 and are
spaced at equal angular intervals of 2𝜋/3
radians, beginning with the root whose
argument is 𝜋/6.

19. Quadratic Formula
The quadratic formula is perfectly valid when the coefficients 𝑎 ≠ 0, 𝑏, and 𝑐 of a
quadratic polynomial equation 𝑎𝑧2
+ 𝑏𝑧 + 𝑐 = 0 are complex numbers. Although
the formula can be obtained in exactly the same manner as for a quadratic
polynomial 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0 where the coefficients 𝑎 ≠ 0, 𝑏, and 𝑐 are real.
However, we choose to write the result as:
𝑧 =
−𝑏 + 𝑏2
− 4𝑎𝑐 1/2
2𝑎
. (∗)
Notice that the numerator of the right-hand side of (∗) looks a little different than
the traditional −𝑏 ± 𝑏2 − 4𝑎𝑐 . Keep in mind that when 𝑏2
− 4𝑎𝑐 ≠ 0, the
expression 𝑏2
− 4𝑎𝑐 1/2
represents the set of two square roots of the complex
number𝑏2
− 4𝑎𝑐. Thus, (∗) gives two complex solutions.

20. Example
Solve the quadratic equation 𝑧2
+ 1 − 𝑖 𝑧 − 3𝑖 = 0.
Solution:
Using quadratic formula with 𝑎 = 1, 𝑏 = 1 − 𝑖, and 𝑐 = −3𝑖, we have:
𝑧 =
− 1 − 𝑖 + 1 − 𝑖 2
− 4(−3𝑖) 1/2
2
=
−1 + 𝑖 + 10𝑖 1/2
2
.
To compute 10𝑖 1/2
we will use the procedure for finding roots of any complex
numbers. For the present case 𝑧 = 10𝑖 with 𝑟 = 10 and 𝜃 = 𝜋/2. We are required
to determine two square roots of 10𝑖. Here 𝑛 = 2 and 𝑘 = 0,1. Thus, the two square
roots of 10𝑖 are given as:
𝜔0 = 10 cos
𝜋
4
+ 𝑖 sin
𝜋
4
= 5 1 + 𝑖 ,
and
𝜔1 = 10 cos
5𝜋
4
+ 𝑖 sin
5𝜋
4
= − 5 1 + 𝑖 .

21. Example
Thus,
𝑧 =
−1 + 𝑖 + 10𝑖 1/2
2
.
possess two root which are given as:
𝑧1 =
1
2
−1 + 𝑖 + 5 + 𝑖 5 =
5 − 1
2
+ 𝑖
5 + 1
2
,
and
𝑧2 =
1
2
−1 + 𝑖 − 5 − 𝑖 5 =
− 5 + 1
2
− 𝑖
5 − 1
2
.

22. Factoring a Quadratic Polynomial
By finding all the roots of a polynomial equation we can factor the polynomial
completely. This statement follows as a corollary to an important theorem that will
be proved later. For the present, note that if 𝑧1 and 𝑧2 are the roots defined by (∗),
i.e.,
𝑧 =
−𝑏 + 𝑏2
− 4𝑎𝑐 1/2
2𝑎
. (∗)
Then a quadratic polynomial 𝑎𝑧2 + 𝑏𝑧 + 𝑐 factors as:
𝑎𝑧2 + 𝑏𝑧 + 𝑐 = 𝑎 𝑧 − 𝑧1 𝑧 − 𝑧2 . (∗∗)

23. Example
Factorize the quadratic polynomial 𝑧2
+ 1 − 𝑖 𝑧 − 3𝑖 .
Solution:
Using equation (**), we have
𝑧2
+ 1 − 𝑖 𝑧 − 3𝑖 = 𝑧 − 𝑧1 𝑧 − 𝑧2
= 𝑧 −
5 − 1
2
− 𝑖
5 + 1
2
𝑧 +
5 + 1
2
+ 𝑖
5 + 1
2
.

24. Practice Questions
1. Find the three cube roots of 𝑧 = −1 + 𝑖.
2. Find the squares of all the cube roots of 𝑧 = − 𝑖.
3. Find the four fourth roots of 𝑧 = −2 3 + 2𝑖.
4. Find the squares of all the 5th roots of 𝑧 =
1
2
+
3
2
𝑖.
5. Find the six 6th roots of (i) −1 and (ii) 1 + 𝑖.

25. Book: A First Course in Complex Analysis with Applications by Dennis G.
Zill and Patrick D. Shanahan.
Chapter: 1
Exercise: 1.4
Q # 1 – 20, 25 – 26.
Exercise: 1.6
Q # 1 –12.
Practice Questions

26. Complex Magic
▪ Every complex number can be transformed into polar form. For 𝑧 = 𝑖 , we obtain:
𝑖 = 𝑒𝑖𝜋/2
.
Thus, in reality 𝑖 correspond to rotation by
𝜋
2
radians (90 degrees). This is the reason
imaginary part is always sketched on the 𝑦 −axis as it is at perpendicular to 𝑥 −axis.
▪ Suppose we have two complex numbers expressed in exponential form 𝑧1 = 𝑟1𝑒𝑖𝜃1
and 𝑧2 = 𝑟2𝑒𝑖𝜃2, then their product is defined as:
𝑧1𝑧2 = 𝑟1𝑒𝑖𝜃1 𝑟2𝑒𝑖𝜃2 = 𝑟1𝑟2 × 𝑒𝑖(𝜃1+𝜃2)
.
r e - s c a l i n g
▪ Therefore, we conclude that a complex product encodes an information of two real
physical operations re-scaling and rotation. We can think on similar lines about other
complex operations e.g., addition, division, complexconjugation.
ro t a t i o n

27. Complex Magic
Let us explore the complex domain in moredetail. Consider a real quadratic equation
𝑥2
− 1 = 0.
Weknow that the solutions of this equationare𝑥 = ±1, which aretwo points on the
real line.

28. Complex Magic
▪ Now consider the complex equation
𝑧2
− 1 = 0.
▪ How many solutions does this complex equation has?
▪ Of course, two: 𝑧 = ±1, two points on the complex plane. If the answers are same, then
we might believe thatboth equations are same !!!.
▪ It is certainly nottrue.
▪ The beauty of secondequation can only be seenif wesubstitute 𝑧 = 𝑥 + 𝑖𝑦, into the
equation,i.e,
(𝑥 + 𝑖𝑦)2
− 1 = 0
𝑥2
− 𝑦2
+ 2𝑖𝑥𝑦 − 1 = 0
𝑥2
− 𝑦2
− 1 + 𝑖2𝑥𝑦 = 0
which gives rise to two equations:
𝑥2
− 𝑦2
= 1 and 2𝑥𝑦 = 0,
▪ The first is an equation of a hyperbola and the other is an equation of 𝑥 − or𝑦 −axis.

29. Complex Magic
Therefore, we can conclude that the two solutions 𝑧 = ±1, are basically two points on
the intersection ofhyperbolas and 𝑥 − axis.

30. Some important facts about complex numbers
1. Space of allcomplex numbers ℂ, is a vectorspace.
2.
1
𝑧
=
ҧ
𝑧
𝑧 2 , where 𝑧 = 𝑥 + 𝑖𝑦.
3. 𝑧1 + 𝑧2 ≤ 𝑧1 + 𝑧2 . (triangular inequality)
4. An 𝑛th degree complex polynomial equation has n complex roots. On the other
hand, an 𝑛th degree real polynomial equation may or may not have 𝑛 real roots.
5. If 𝑧 = 𝑥 + 𝑖𝑦 we define 𝑒𝑧 = 𝑒𝑥+𝑖𝑦 = 𝑒𝑥 cos 𝑦 + 𝑖 sin 𝑦 .
6. 𝑒𝑧
is never zero.
7. 𝑒𝑧
= 𝑒𝑥
, where 𝑧 = 𝑥 + 𝑖𝑦.