de Moivres Theorem
de Moivres Theorem
de Moivres Theorem
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
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.
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
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 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.