USE DE MOIVRES THEOREM TO FIND THE INDICATED POWER OF COMPLEX NUMBER

Use DeMoivre's Theorem to find the indicated power of the complex number. Write answers in rectangular form.

Problem 1 :

[4(cos 15° + i sin 15°)]³

Solution:

Let z = [4(cos 15° + i sin 15°)]³

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 3, r = 4 and θ = π/12

By using the calculator, we get

Problem 2 :       

[2(cos 10° + i sin 10°)]³

Solution:

Let z = [2(cos 10° + i sin 10°)]³

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 3, r = 2 and θ = π/18

Problem 3 :       

[2(cos 80° + i sin 80°)]³

Solution:

Let z = [2(cos 80° + i sin 80°)]³

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 3, r = 2 and θ = 4π/9

Problem 4 :

[2(cos 40° + i sin 40°)]³

Solution:

Let z = [2(cos 40° + i sin 40°)]³

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 3, r = 2 and θ = 2π/9

Problem 5 :

Solution:

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 6, r = 1/2 and θ = π/12

Problem 6 :

Solution:

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 5, r = 1/2 and θ = π/10

Problem 7 :

Solution:

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 4, r = √2 and θ = 5π/6

Problem 8 :

Solution:

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 6, r = √3 and θ = 5π/18

Problem 9 :

(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

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 5, r = √2 and θ = π/4

Problem 10 :

(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 and negative, z lies in the fourth quadrant.

So, the principal value θ = -π/4

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

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 5, r = √2 and θ = -π/4

Problem 11 :

(√3 - i)⁶

Solution:

Given, standard form z = (√3 - i)

The polar form of the complex number z is 

(√3 - i) = r cos θ + i sin θ --- (1)

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

So, the principal value θ = -π/6

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

Using the De Moivre's formula,

zⁿ = rⁿ(cos nθ + i sin nθ)

Here n = 6, r = 2 and θ = -π/6

Problem 12 :

(√2 - i)⁴

Solution:

Given, standard form z = (√2 - i)

The polar form of the complex number z is 

(√2 - i) = r cos θ + i sin θ --- (1)