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de Moivres Theorem

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De Moivre’s

Theorem and
its
Applications
Dr. Hina Dutt
hina.dutt@seecs.edu.pk
SEECS-NUST
Advanced
Engineering
Mathematics (10th
• Chapter: 13
Edition) by Ervin • Sections: 13.2
Kreyszig

A First Course in
Complex Analysis
with Applications by
• Chapter: 1
Dennis G. Zill and • Sections: 1.3, 1.4, 1.6
Patrick D. Shanahan.
De Moivre’s Theorem

o De Moivre's theorem, named after the French


mathematician Abraham de Moivre is used to find the
roots of a complex number for any power n, given that 𝑛 is
an integer.
o De Moivre's theorem can be derived from Euler's equation,
and is important because it connects trigonometry to complex
numbers.
De Moivre’s Theorem

For any complex number 𝑧 and any integer 𝑛, the


following is true:
If 𝑧 = 𝑟 cos 𝜃 + 𝑖 sin 𝜃 ,
Then 𝑧 𝑛 = 𝑟 𝑛 (cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃)
Example 1
3
Expand − 3 − 𝑖 .
Example 1
3
We can expand − 3 − 𝑖 in two ways.

Method 1:
Using Binomial Theorem
3 3 2
2 3
− 3−𝑖 = − 3 +3 − 3 −𝑖 + 3 − 3 −𝑖 + −𝑖
= −8𝑖
Example 1

Method 2:
Using De Moivre’s Theorem
3
3 7𝜋 7𝜋 7𝜋 7𝜋
− 3−𝑖 = 2 cos + 𝑖 sin = 23 cos 3 × + 𝑖 sin 3 × = −8𝑖
6 6 6 6
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.


Root of a Complex Number

Recall from algebra that – 2 and 2 are said to be square roots of the
number 4 because −2 2 = 4 and 2 2 = 4. In other words, the two
square roots of 4 are distinct solutions of the equation 𝜔2 = 4.
In like manner, we say 𝜔 = 3 is a cube root of 27 since 𝜔3 = 33 = 27.
Root of a Complex Number
A number 𝑤 is an 𝒏𝒕𝒉 root of a nonzero complex number 𝑧 if
𝜔𝑛 = 𝑧, where 𝑛 is a positive integer.

1 1 1 1
For example, 𝜔1 = 2 + 2𝑖 and 𝜔2 = − 2 − 2𝑖 are two
2 2 2 2
complex roots of a complex number 𝑧 = 𝑖 because 𝑤12 = 𝑖 and 𝜔22 = 𝑖.
nth Root of a Complex Number
The 𝑛 𝑛𝑡ℎ root of a nonzero complex number 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃) are given
by
1 𝜃 + 2𝑘𝜋 𝜃 + 2𝑘𝜋
𝜔𝑘 = 𝑟 𝑛 cos + 𝑖 sin ; 𝑘 = 0, 1, 2, … , 𝑛 − 1
𝑛 𝑛

1
These 𝑛 values lie on a circle of radius 𝑟 with center at the origin and
𝑛

constitute the vertices of a regular polygon of 𝑛 sides.


Principal nth Root of a Complex Number

The unique root of a complex number 𝑧 obtained by using the


principal value of arg(𝑧) with 𝑘 = 0 is naturally referred to as the
principal 𝒏𝒕𝒉 root of 𝜔.
Example 2
Find the cube root of 𝑧 = 𝑖.
Solution:
We have to solve 𝜔3 = 𝑖
𝜋
With 𝑟 = 1 and arg 𝑖 = , we obtain
2
𝜋 𝜋
1 + 2𝑘𝜋 + 2𝑘𝜋
𝜔𝑘 = 13 cos 2 + 𝑖 sin 2 , 𝑘 = 0, 1, 2.
3 3
Hence the roots are
𝜋 𝜋 3 1
𝑘 = 0, 𝜔0 = cos + 𝑖 sin = + 𝑖
6 6 2 2
5𝜋 5𝜋 3 1
𝑘 = 1, 𝜔1 = cos + 𝑖 sin =− + 𝑖
6 6 2 2
3𝜋 3𝜋
𝑘 = 2, 𝜔2 = cos + 𝑖 sin = −𝑖
2 2
Example 2

𝜋
Notice 𝐴𝑟𝑔 𝑧 = , we see that
2
3 1
𝜔0 = + 𝑖 is the principal root of 𝑖.
2 2
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.
Quadratic Formula
The quadratic formula is valid when the coefficients 𝑎 ≠ 0, 𝑏,
and 𝑐 of a quadratic polynomial equation 𝑎𝑧 2 + 𝑏𝑧 + 𝑐 = 0 are
complex numbers.

−𝑏 + 𝑏 2 − 4𝑎𝑐 1/2
𝑧=
2𝑎
Factoring a Quadratic Polynomial
If 𝑧1 and 𝑧2 are the roots of a quadratic equation 𝑎𝑧 2 + 𝑏𝑧 + 𝑐 = 0
then the quadratic polynomial 𝑎𝑧 2 + 𝑏𝑧 + 𝑐 can be factored as

𝑎𝑧 2 + 𝑏𝑧 + 𝑐 = 𝑎(𝑧 − 𝑧1 )(𝑧 − 𝑧2 )
Questions

a. Find the four fourth roots of 𝑧 = 1 + 𝑖.


b. Find the squares of all cube roots of 𝑧 = −𝑖.
c. Solve the equation 𝑧 2 + 1 + 𝑖 𝑧 + 5𝑖 = 0.
d. Factorize the quadratic polynomial 𝑧 2 + 1 + 𝑖 𝑧 + 5𝑖.
Practice Questions

A First Course in
Complex Analysis • Chapter: 1
with Applications • Exercise: 1.3 Questions: 25-32
by Dennis G. Zill • Exercise: 1.4 Questions: 1-19
and Patrick D. • Exercise: 1.6 Questions: 1-12
Shanahan.

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