Solved Example Problems on de Moivre’s Theorem

Example 2.28

If z = (cosθ + i sinθ ) , show that zn + 1/ zn = 2 cos nθ and zn â€“ [1/ zn] = 2i sin nθ .

Solution

Let z = (cosθ + i sinθ ) . 

By de Moivre’s theorem ,

zn = (cosθ + i sinθ )n = cos nθ + i sin nθ 

Example 2.29

Similarly, 

Solution

Example 2.30

Solution

Example 2.31

Simplify 

(i) (1+ i)18

(ii) (-√3 + 3i)31 .

Solution

(i) (1+ i)18

Let 1+ i = r (cosθ + i sinθ ) . Then, we get

(ii) (-√3 + 3i)31 .

Let -√3 + 3i = r (cosθ + i sinθ ) . Then, we get

Raising power 31 on both sides,

The nth  roots of unity

Example 2.32

Find the cube roots of unity. 

Solution

We have to find 11/3 . Let z = 11/3 then z3 = 1.

In polar form, the equation z3 = 1 can be written as

z3 = cos(0 + 2kÏ€) + i sin(0 + 2kÏ€) = ei2kÏ€ , k = 0, 1, 2,... 

Example 2.33

Find the fourth roots of unity.

Solution

We have to find 11/4. Let z =11/4 . Then z4 = 1 .

In polar form, the equation z4 = 1 can be written as

z4 = cos (0 + 2kÏ€ ) + i sin (0 + 2kÏ€ ) = ei2kÏ€ , k = 0, 1, 2,...

Note

(i) In this chapter the letter ω is used for nth roots of unity. Therefore the value of ω is depending on n as shown in following table.

(ii) The complex number zeiθ is a rotation of z by θ radians in the counter clockwise direction about the origin.

Example 2.34

Solve the equation z3 + 8i = 0 , where z âˆˆ C. 

Solution

Let z3 + 8i = 0 . Then, we get

z3 = -8i

Example 2.35

Find all cube roots of √3 + i 

Solution

We have to find (√3 + i)1/3. Let z3 = √3 + i = r (cosθ + i sinθ )

Example 2.36

Suppose z1 , z2, and z3 are the vertices of an equilateral triangle inscribed in the circle |z| = 2. If z1 = 1+ i√3 , then find z2 and z3.

Solution

|z| = 2 represents the circle with centre (0, 0) and radius 2.

Let A, B, and C be the vertices of the given triangle. Since the vertices z1 , z2 , and z3 form an equilateral triangle inscribed in the circle |z| = 2 , the sides of this triangle AB, BC, and CA subtend 2Ï€/3 radians (120 degree) at the origin (circumcenter of the triangle).

(The complex number z eiθ is a rotation of z by θ radians in the counter clockwise direction about the origin.)

Therefore, we can obtain z2 and z3 by the rotation of z1 by 2Ï€/3 and 4 Ï€/3 respectively.

Given that 

Therefore, z2 = -2, and z3 = 1- i√3.