Solutions — Polar Form and De Moivre’s Theorem

Q1 — Converting to Polar Form

To convert z = a + bi to polar form: compute r = √(a²+b²), then find θ by identifying the quadrant.

(a) z = √3 + i: a = √3 > 0, b = 1 > 0 → Q1.

r = √(3 + 1) = √4 = 2. tanθ = 1/√3 → θ = π/6.

z = 2 cis(π/6)

(b) z = −1 + i: a = −1 < 0, b = 1 > 0 → Q2.

r = √(1 + 1) = √2. Reference angle: arctan(1/1) = π/4. θ = π − π/4 = 3π/4.

z = √2 cis(3π/4)

(c) z = −2: this lies on the negative real axis.

r = 2. θ = π.

z = 2 cis(π)

(d) z = −i = 0 − i: this lies on the negative imaginary axis.

r = 1. θ = −π/2.

z = cis(−π/2)

Q2 — Converting from Polar to Cartesian

Use z = r cisθ = r cosθ + i r sinθ.

(a) 4 cis(0): a = 4 cos(0) = 4(1) = 4, b = 4 sin(0) = 0. z = 4

(b) 2 cis(π/2): a = 2 cos(π/2) = 2(0) = 0, b = 2 sin(π/2) = 2(1) = 2. z = 2i

(c) 3 cis(2π/3): a = 3 cos(2π/3) = 3(−1/2) = −3/2, b = 3 sin(2π/3) = 3(√3/2) = 3√3/2.

z = −3/2 + (3√3/2)i

(d) 6 cis(−π/6): a = 6 cos(−π/6) = 6(√3/2) = 3√3, b = 6 sin(−π/6) = 6(−1/2) = −3.

z = 3√3 − 3i

Q3 — Multiplying and Dividing in Polar Form

z&sub1; = 3 cis(π/4), z&sub2; = 2 cis(π/6).

(a) Multiplication rule: multiply moduli, add arguments.

|z&sub1;z&sub2;| = 3 × 2 = 6

arg(z&sub1;z&sub2;) = π/4 + π/6 = 3π/12 + 2π/12 = 5π/12

z&sub1;z&sub2; = 6 cis(5π/12)

(b) Division rule: divide moduli, subtract arguments.

|z&sub1;/z&sub2;| = 3/2

arg(z&sub1;/z&sub2;) = π/4 − π/6 = 3π/12 − 2π/12 = π/12

z&sub1;/z&sub2; = (3/2) cis(π/12)

Q4 — Applying De Moivre’s Theorem (Basic)

De Moivre: (r cisθ)ⁿ = rⁿ cis(nθ).

(a) [2 cis(π/3)]³ = 2³ cis(3 × π/3) = 8 cis(π)

cis(π) = cosπ + i sinπ = −1 + 0i = −1

Answer: −8

(b) [cis(π/6)]¹² = 1¹² cis(12 × π/6) = cis(2π)

cis(2π) = cos(2π) + i sin(2π) = 1 + 0i = 1

Answer: 1

(c) [√2 cis(π/4)]⁴ = (√2)⁴ cis(4 × π/4) = (2^1/2)⁴ cis(π) = 2² cis(π)

= 4 × (−1) = −4

Q5 — Powers of Complex Numbers

Step 1 — Convert to polar:

1 + i: a = 1, b = 1 → Q1.

r = √(1+1) = √2 θ = arctan(1/1) = π/4

So 1 + i = √2 cis(π/4).

Step 2 — Apply De Moivre:

(1 + i)⁶ = (√2)⁶ cis(6 × π/4)

(√2)⁶ = (2^1/2)⁶ = 2³ = 8

6 × π/4 = 6π/4 = 3π/2

= 8 cis(3π/2)

Step 3 — Convert back:

8 cis(3π/2) = 8[cos(3π/2) + i sin(3π/2)] = 8[0 + i(−1)] = −8i

Q6 — Square Roots in Polar Form

For w² = z = r cisθ, the square roots are w = √r cis((θ + 2kπ)/2), k = 0, 1.

Here z = 4 cis(π/3), so r = 4, θ = π/3.

k = 0: w = √4 cis((π/3 + 0)/2) = 2 cis(π/6)

= 2[cos(π/6) + i sin(π/6)] = 2[√3/2 + i/2] = √3 + i (as a check)

k = 1: w = 2 cis((π/3 + 2π)/2) = 2 cis((7π/3)/2) = 2 cis(7π/6)

Principal argument: 7π/6 > π, so subtract 2π: 7π/6 − 2π = −5π/6

= 2 cis(−5π/6)

Note: the two roots are 2 cis(π/6) and 2 cis(π/6 + π) = 2 cis(7π/6) — they differ by π in argument, confirming they are negatives of each other.

Q7 — Geometric Interpretation

Recall: multiplying two complex numbers multiplies their moduli and adds their arguments.

(a) i = 1 cis(π/2). Multiplying z by i:

• Modulus: |z| × 1 = |z| (unchanged)

• Argument: arg(z) + π/2

Result: anticlockwise rotation by 90° about the origin.

(b) −1 = 1 cis(π). Multiplying z by −1:

• Modulus unchanged • Argument increases by π

Result: rotation by 180° (reflection through the origin).

(c) 2 cis(π/3): modulus = 2, argument = π/3. Multiplying z by this:

• Modulus: |z| × 2 (dilation by factor 2)

• Argument: arg(z) + π/3 (rotation anticlockwise by 60°)

Result: dilation by factor 2 and anticlockwise rotation by 60°.

Q8 — Modulus and Argument of a Product

Polar form of z = 1 − i:

|z| = √(1² + (−1)²) = √2

Q4 (a > 0, b < 0): θ = −arctan(1/1) = −π/4

Polar form of w = √3 + i:

|w| = √(3 + 1) = 2

Q1: θ = arctan(1/√3) = π/6

Product:

|zw| = |z| × |w| = √2 × 2 = 2√2

arg(zw) = arg(z) + arg(w) = −π/4 + π/6 = −3π/12 + 2π/12 = −π/12

Q9 — Cube Roots of Unity

We solve z³ = 1 using polar form.

Write 1 = 1 cis(0 + 2kπ) for k = 0, 1, 2.

The cube roots are z = 1^1/3 cis(2kπ/3) = cis(2kπ/3).

k = 0: cis(0) = 1 + 0i = 1

k = 1: cis(2π/3) = cos(2π/3) + i sin(2π/3)

cos(2π/3) = −1/2, sin(2π/3) = √3/2

= −1/2 + (√3/2)i

k = 2: cis(4π/3) = cos(4π/3) + i sin(4π/3)

cos(4π/3) = −1/2, sin(4π/3) = −√3/2

= −1/2 − (√3/2)i

These three roots lie on the unit circle, equally spaced at 120° apart. Note that the two non-real roots are conjugates of each other, consistent with the conjugate root theorem.

Q10 — Using De Moivre to Derive a Trigonometric Identity

Strategy: Compute (cosθ + i sinθ)² two ways and compare real and imaginary parts.

De Moivre’s side: (cosθ + i sinθ)² = cos(2θ) + i sin(2θ)

Algebraic expansion:

(cosθ + i sinθ)² = cos²θ + 2i sinθ cosθ + i² sin²θ

= cos²θ + 2i sinθ cosθ − sin²θ

= (cos²θ − sin²θ) + i(2 sinθ cosθ)

Equating parts:

Real parts: cos(2θ) = cos²θ − sin²θ ✓

Imaginary parts: sin(2θ) = 2 sinθ cosθ ✓

This approach generalises: using (cosθ + i sinθ)³ similarly derives the triple angle formulas.