FIND THE INDICATED POWER OF A COMPLEX NUMBER

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

z⁵  =  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)  

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,

z⁵  =  [√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)  

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,

z³  =  [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  =  2 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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