Formula Name	Formula
Polar Form of a Complex Number	z = r (cosθ + i sinθ) where, r = ∣z∣ (modulus) and θ = arg(z) (argument)
Euler’s Formula	eiθ = cosθ + i sinθ
De Moivre’s Theorem	[r (cosθ + i sinθ)] n = rn [cos(nθ) +i sin(nθ)]
Roots of a Complex Number	For a complex number z = r (cosθ + i sinθ) the nth roots are: n√z = n√r [cos (nθ +2kπ​) + i sin (nθ +2kπ​)] where, k = 0, 1, 2, …, n-1

De Moivres Theorem Practice Problems

Problem 1: Find (1 + i)⁴ using De Moivre’s Theorem.

Solution:

Convert (1 + i ) to polar form.

r = √ (1² + 1²) = √2

θ = tan^-1(1/1) = tan^-1(1) = π/4

Apply De Moivre’s Theorem:

(1 + i)⁴ = [√2 (cos (π/4) + i sin (π/4)]⁴

(1 + i)⁴ = (√2)⁴ [cos {4 × (π/4)} + i sin {4 × (π/4)}]

(1 + i)⁴ = 4 [cos π + i sin π]

(1 + i)⁴ = 4 (-1 + 0i) = -4

Answer: -4

Question 2: Compute [2(cos (π/6) + i sin (π/6)]³.

Solution:

Apply De Moivre’s Theorem:

[2(cos (π/6) + i sin (π/6)]³ = 2³ [cos {3 × (π/6)} + i sin {3 × (π/6)}]

[2(cos (π/6) + i sin (π/6)]³ = 8 [cos (π/2) + i sin (π/2)]

[2(cos (π/6) + i sin (π/6)]³ = 8 (0 + i)

[2(cos (π/6) + i sin (π/6)]³ = 8i

Answer: 8i

Question 3: Find the cube root of 8 using De Moivre’s Theorem.

Solution:

Express 8 as a complex number in polar form:

8 = 8 (cos 0 + i sin 0)

The cube roots are given by:

∛8 = ∛8 [cos {(0 + 2kπ)/3} + i sin {(0 + 2kπ)/3}]

∛8 = 2 [cos (2kπ/3) + i sin (2kπ/3)]

For k = 0

∛8 = 2 (cos 0 + i sin 0) = 2

For k = 1

∛8 = 2 [cos (2kπ/3) + i sin (2kπ/3)] = 2 [cos (2π/3) + i sin (2π/3)]

∛8 = 2 [(-1/2) + i (√3/2)]

∛8 = -1 + i√3

For k = 2

∛8 = 2 [cos (2kπ/3) + i sin (2kπ/3)] = 2 [cos (4π/3) + i sin (4π/3)]

∛8 = 2 [(-1/2) – i (√3/2)] = -1 – i√3

Answer: 2, (-1 + i√3), (-1 – i√3)

Question 4: Use De Moivre’s Theorem to find (1 – i)⁵.

Solution:

Convert (1 – i) to polar form

r = √ (1² + (-1)²) = √2

θ = tan⁻¹(2/2​) = tan⁻¹(1) = π /4​

Apply De Moivre’s Theorem

(1 – i)⁵ = [√2(cos (π/4) + i sin (π/4))]⁵

(1 – i)⁵ = (√2)⁵ [cos 5(π/4) + i sin 5(π/4)]

(1 – i)⁵ = 4√2[cos (5π/4) + i sin (5π/4)]

(1 – i)⁵ = 4√2[ (-1/√2) – i(1/√2)]

(1 – i)⁵ = 4 (-1 -i )

(1 – i)⁵ = -4 – 4i

Answer: -4 – 4i

Question 5: Calculate (2 + 2i)³ using De Moivre’s Theorem.

Solution:

Convert (2 + 2i) to polar form

r = √ (2² + (2)²) = 2√2

θ = tan⁻¹(2/2​) = tan⁻¹(1) = π /4​

Apply De Moivre’s Theorem

(2 – 2i)³ = [2√2(cos (π/4) + i sin (π/4))]³

(2 – 2i)³ = (2√2)³ [cos 3(π/4) + i sin 3(π/4)]

(2 – 2i)³ = 16√2[cos (3π/4) + i sin (3π/4)]

(2 – 2i)³ = 16√2[ (-1/√2) – i(1/√2)]

(2 – 2i)³ = 16 (-1 + i)

(2 – 2i)³ = -16 + 16i

Answer: -16 + 16i

Question 6: Determine the fourth roots of 16 using De Moivre’s Theorem.

Solution:

Express 16 as a complex number in polar form:

16 = 16 (cos 0 + i sin 0)

Fourth roots are given by:

∜16 = ∜16 [cos {(0 + 2kπ)/4} + i sin {(0 + 2kπ)/4}]

∜16 = 2 [cos (2kπ/4) + i sin (2kπ/4)]

For k = 0

∜16 = 2 (cos 0 + i sin 0) = 2

For k = 1

∜16 = 2 [cos (2kπ/4) + i sin (2kπ/4)] = 2 [cos (π/2) + i sin (π/2)]

∜16 = 2 [(0) + i (1)]

∜16 = 2(0 + i) = 2i

For k = 2

∜16 = 2 [cos (2kπ/4) + i sin (2kπ/4)] = 2 [cos (π) + i sin (π)]

∜16 = 2(-1 + 0i) = -2

For k = 3

∜16 = 2 [cos (2kπ/4) + i sin (2kπ/4)] = 2 [cos (3π/2) + i sin (3π/2)]

∜16 = 2 [0 – i] = – 2i

Answer: 2, 2i, -2, -2i

Question 7: Calculate [3(cos (π/4) + i sin (π/4))]²

Solution:

Apply De Moivre’s Theorem

[3(cos (π/4) + i sin (π/4))]² = 3²[cos (2π/4) + i sin (2π/4)]

[3(cos (π/4) + i sin (π/4))]² = 9 [cos (π/2) + i sin (π/2)]

[3(cos (π/4) + i sin (π/4))]² = 9(0 + i)

[3(cos (π/4) + i sin (π/4))]² = 9i

Answer: 9i

Question 8: Use De Moivre’s Theorem to find (1 + √3i)⁶.

Solution:

Convert (1 + √3i)⁶ to polar form

r = √ (1² + (√3)²) = √4 = 2

θ = tan⁻¹(√3/1​) = π / 3

Apply De Moivre’s Theorem

(1 + √3i)⁶ = [2(cos (π/3) + i sin (π/3))]⁶

(1 + √3i)⁶ = (2)⁶ [cos 6(π/3) + i sin 6(π/3)]

(1 + √3i)⁶ = 64[cos (2π) + i sin (2π)]

(1 + √3i)⁶ = 64(1 + 0i)

(1 + √3i)⁶ = 64

Answer: 64

Question 9: Compute (1 – √3i)³ using De Moivre’s Theorem.

Solution:

Convert (1 – √3i) to polar form

r = √ (1² + (√3)²) = √4 = 2

θ = tan⁻¹(-√3/1​) = – π / 3

Apply De Moivre’s Theorem

(1 – √3i)³ = [2(cos (-π/3) + i sin (-π/3))]³

(1 – √3i)³ = (2)³ [cos 3(-π/3) + i sin 3(-π/3)]

(1 – √3i)³ = 8[cos (-π) + i sin (-π)]

(1 – √3i)³ = 8(-1 – 0i)

(1 – √3i)³ = -8

Answer: -8

Question 10: Determine (√3 + i)⁴ using De Moivre’s Theorem.

Solution:

Convert (√3 + i) to polar form

r = √ ((√3)² + 1²) = √4 = 2

θ = tan⁻¹(1/√3​) = π / 6

Apply De Moivre’s Theorem

(√3 + i)⁴ = [2(cos (π/6) + i sin (π/6))]⁴

(√3 + i)⁴ = (2)⁴ [cos 4(π/6) + i sin 4(π/6)]

(√3 + i)⁴ = 16[cos (2π/3) + i sin (2π/3)]

(√3 + i)⁴ = 16[(-1/2) + (√3​/2)i]

(√3 + i)⁴ = -8 + 8√3i

Answer: -8 + 8√3i

De Moivre’s Theorem Practice Problems

Problem 1. Use De Moivre’s Theorem to find (2 + 3i)³.

Problem 2. Compute [8(cos (π/2) + i sin (π/2))]³ using De Moivre’s Theorem.

Problem 3. Determine the fourth roots of 27 using De Moivre’s Theorem.

Problem 4. Find (5 + i)⁵ using De Moivre’s Theorem.

Problem 5. Calculate [5(cos (π/3) + i sin (π/3))]⁴.

Problem 6. Use De Moivre’s Theorem to find (1 + ∛i)⁶.

Problem 7. Determine the square roots of -9 using De Moivre’s Theorem.

Problem 8. Compute [4(cos (π/2) + i sin (π/2))]³ using De Moivre’s Theorem.

Problem 9. Find (2 – i)⁴ using De Moivre’s Theorem.

Problem 10. Calculate [3(cos (π/6) + i sin (π/6))]⁵

Frequently Asked Questions

What is the General Formula for De Moivre’s Theorem?

The general formula for De Moivre’s Theorem is,

[r (cosθ + i sinθ)] ⁿ = rⁿ [cos(nθ) +i sin(nθ)]

How is De Moivre’s Theorem Useful?

De Moivre’s Theorem is useful for:

• Raising complex numbers to a power.
• Finding the roots of complex numbers.
• Simplifying expressions involving trigonometric functions.

What are the Limitations of De Moivre’s Theorem?

De Moivre’s Theorem is primarily used for integer powers and roots of complex numbers. It doesn’t directly apply to non-integer exponents or to more general complex functions without further modifications.

How Do You Convert a Complex Number to Polar Form?

A complex number a + bi can be converted to polar form r(cos θ + i sinθ)

where, r = √ (a² + b²) and θ = tan⁻¹(-b/a​)