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

Example 6 :

(-1 - 6i)³

Solution :

Given, standard form z  =  (-1 - 6i)³

Comparing with (a - b)³, we get a³ - 3a²b + 3ab² - b³

a = -1 and b = 6i

= (-1)³ - 3(-1)²(6i) + 3(-1)(6i)² - (6i)³

= -1 - 18i - 3(36i²) - 216i³

= -1 - 18i - 3(36(-1)) - 216i² i

= -1 - 18i + 108 - 216(-1) i

= -1 - 18i + 108 + 216i

= 107 + 198i

Example 6 :

(3 – 2i)(5 + 4i) – (3 – 4i)²

Solution :

(3 – 2i)(5 + 4i) – (3 – 4i)²

(3 – 2i)(5 + 4i) = 3(5) + 3(4i) + (-2i)(5) - 2i(4i)

= 15 + 12i - 10i - 8i²

= 15 + 2i - 8(-1)

= 15 + 2i + 8

= 23 + 2i

Example 7 :

(1 - 2i)² - (1 + 2i)²

Solution :

(1 - 2i)² - (1 + 2i)²

(a - b)² = a² - 2ab + b²

(1 + 2i)² = 1² - 2(1)(2i) + (2i)²

= 1 - 4i + 4i²

= 1 - 4i - 4

= -3 - 4i

(1 + 2i)² = 1² + 2(1)(2i) + (2i)²

= 1 + 4i + 4i²

= 1 + 4i - 4

= -3 + 4i

(1 - 2i)² - (1 + 2i)² = -3 - 4i - (-3 + 4i)

= -3 - 4i + 3 - 4i

= -8i

(1 - 2i)² - (1 + 2i)² = -8i

Example 7 :

(2 + 3i)² - (2 - 3i)²

Solution :

(2 + 3i)² - (2 - 3i)²

(a - b)² = a² - 2ab + b²

(2 + 3i)² = 2² + 2(2)(3i) + (3i)²

= 4 + 12i + 9i²

= 4 + 12i - 9

= -5 + 12i

(2 - 3i)² = 2² - 2(2)(3i) + (3i)²

= 4 - 12i + 9i²

= 4 - 12i - 9

= -5 - 12i

(2 + 3i)² - (2 - 3i)² = -5 + 12i - (-5 - 12i)

= -5 + 12i + 5 + 12i

= 24i

(2 + 3i)² - (2 - 3i)² = 24i

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