**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,

z^{n} = [r(cos θ + i sin θ)]^{n}

z^{n }= r^{n}(cos nθ + i sin nθ)

For all positive integers n.

**Example 1 :**

(cos π/4 + i sin π/4)^{3}

**Solution :**

Given, z^{3} = (cos π/4 + i sin π/4)^{3}

Using the De Moivre's formula :

z^{n}_{ }= r^{n}(cos nθ + i sin nθ )

Here n = 3, r = 1 and θ = π/4

z^{3}_{ }= 1^{3}(cos 3π/4 + i sin 3π/4)

z^{3}_{ }= 1(cos 3π/4 + i sin 3π/4)

By using the calculator, we get

z^{3}_{ }= -√2/2 + i √2/2

**Example 2 :**

[3(cos 3π/2 + i sin 3π/2)]^{5}

**Solution :**

Given, z^{5} = [3(cos 3π/2 + i sin 3π/2)]^{5}

Using the De Moivre's formula :

z^{n}_{ }= r^{n}(cos nθ + i sin nθ )

Here n = 5, r = 3 and θ = 3π/2

z^{5}_{ }= 3^{5}(cos 15π/2 + i sin 15π/2)

By using the calculator, we get

z^{5 }= 243(0 - i)

z^{5}_{ }= 243i

**Example 3 :**

[2(cos 3π/4 + i sin 3π/4)]^{3}

**Solution :**

Given, z^{3} = [2(cos 3π/4 + i sin 3π/4)]^{3}

Using the De Moivre's formula :

z^{n}_{ }= r^{n}(cos nθ + i sin nθ )

Here n = 3, r = 2 and θ = 3π/4

z^{3}_{ }= 2^{3}(cos 9π/4 + i sin 9π/4)

By using the calculator, we get

z^{3}^{ }= 8(√2/2 + i √2/2)

z^{3}^{ }= [(8√2/2) + (i 8√2/2)]

z^{3}^{ }= 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) r = √2 |
Finding α : α = tan 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^{5 }= [√2(cos π/4 + i sin π/4)]^{5}

Using the De Moivre's formula :

z^{n}_{ }= r^{n}(cos nθ + i sin nθ )

Here n = 5, r = √2 and θ = π/4

z^{5}_{ }= (√2)^{5}(cos 5. π/4 + i sin 5 . π/4)

z^{5}_{ }= 4√2(cos 5π/4 + i sin 5π/4)

By using the calculator, we get

z^{5}^{ }= 4√2(-√2/2 - i √2/2)

z^{5}^{ }= [-( 4√2 . √2/2) - (i 4√2. 8√2/2)]

z^{5}^{ }= -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) r = 2 |
Finding α : α = tan 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^{3 }= [2(cos -π/3 + i sin -π/3)]^{3}

Using the De Moivre's formula :

z^{n}_{ }= r^{n}(cos nθ + i sin nθ )

Here n = 3, r = 2 and θ = -π/3

z^{3}_{ }= (2)^{3}[cos (-3π/3) + i sin (-3π/3)]

z^{3}_{ }= 8[cos (-3π/3) + i sin (-3π/3)]

By using the calculator, we get

z^{3}^{ }= 8(-1 - i0)

z^{3}^{ }= -8

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