# FIND THE INDICATED POWER OF A COMPLEX NUMBER

How to find the indicated power of a complex number :

We are using the De Moivre's theorem, to find the indicated power of a complex number.

Definition of De Moivre's theorem :

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.

Let z  =  r(cos θ + i sin θ) be a polar form of a complex number.

According to De Moivre's theorem,

zn  =  [r(cos θ + i sin θ)]n

z=  rn(cos nθ + i sin nθ)

For all positive integers n.

Example 1 :

(cos π/4 + i sin π/4)3

Solution :

Given, z3  =  (cos π/4 + i sin π/4)3

Using the De Moivre's formula :

zn  =  rn(cos nθ + i sin nθ )

Here n  =  3, r  =  1 and θ  =  π/4

z3  =  13(cos 3π/4 + i sin 3π/4)

z3  =  1(cos 3π/4 + i sin 3π/4)

By using the calculator, we get

z3  =  -√2/2 + i √2/2

Example 2 :

[3(cos 3π/2 + i sin 3π/2)]5

Solution :

Given, z5  =  [3(cos 3π/2 + i sin 3π/2)]5

Using the De Moivre's formula :

zn  =  rn(cos nθ + i sin nθ )

Here n  =  5, r  =  3 and θ  =  3π/2

z5  =  35(cos 15π/2 + i sin 15π/2)

By using the calculator, we get

z=  243(0 - i)

z5  =  243i

Example 3 :

[2(cos 3π/4 + i sin 3π/4)]3

Solution :

Given, z3  =  [2(cos 3π/4 + i sin 3π/4)]3

Using the De Moivre's formula :

zn  =  rn(cos nθ + i sin nθ )

Here n  =  3, r  =  2 and θ  =  3π/4

z3  =  23(cos 9π/4 + i sin 9π/4)

By using the calculator, we get

z3  =  8(√2/2 + i √2/2)

z3  =  [(8√2/2) + (i 8√2/2)]

z3  =  4√2 + i 4√2

Example 4 :

(1 + i)5

Solution :

Given, standard form z  =  (1 + i)

The polar form of the complex number z is

(1 + i)  =  r cos θ + i sin θ ----(1)

 Finding r :r  =  √[(1)2 + (1)2]r  =  √2 Finding α :α  =  tan-1(1/1)α  =  π/4

Since the complex number 1 + i is positive, z lies in the first quadrant.

So, the principal value θ  =  π/4

By applying the value of r and θ in equation (1), we get

1 + i  =  √2(cos π/4 + i sin π/4)

So, the polar form of z is √2(cos π/4 + i sin π/4)

Then,

z=  [√2(cos π/4 + i sin π/4)]5

Using the De Moivre's formula :

zn  =  rn(cos nθ + i sin nθ )

Here n  =  5, r  =  √2 and θ  =  π/4

z5  =  (√2)5(cos 5. π/4 + i sin 5 . π/4)

z5  =  4√2(cos 5π/4 + i sin 5π/4)

By using the calculator, we get

z5  =   4√2(-√2/2 - i √2/2)

z5  =  [-( 4√2 . √2/2) - (i  4√2. 8√2/2)]

z5  =  -4 - 4i

Example 5 :

(1 - √3i)3

Solution :

Given, standard form z  =  (1 - √3i)

The polar form of the complex number z is

1 - √3i  =  r cos θ + i sin θ ----(1)

 Finding r :r  =  √[(1)2 + (√3)2]r  =  2 Finding α :α  =  tan-1(√3/1)α  =  π/3

Since the complex number 1 - √3i is positive and negative, z lies in the fourth quadrant.

So, the principal value θ  =  -π/3

By applying the value of r and θ in equation (1), we get

1 - √3i  =  2(cos -π/3 + i sin -π/3)

So, the polar form of z is 2(cos -π/3 + i sin -π/3)

Then,

z=  [2(cos -π/3 + i sin -π/3)]3

Using the De Moivre's formula :

zn  =  rn(cos nθ + i sin nθ )

Here n  =  3, r  =  and θ  =  -π/3

z3  =  (2)3[cos (-3π/3) + i sin (-3π/3)]

z3  =  8[cos (-3π/3) + i sin (-3π/3)]

By using the calculator, we get

z3  =   8(-1 - i0)

z3  =  -8

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