Precalculus — Complex Numbers & Polar Form

The Complex Plane & Polar Form

Plotting complex numbers, converting between rectangular and polar form, multiplying and dividing in polar form, De Moivre's Theorem, and finding nth roots — with worked examples.

The Complex Plane

Every complex number z = a + bi corresponds to a unique point in the complex plane (also called the Argand plane). The plane looks like a standard coordinate grid, but the two axes represent different things:

Horizontal axis

Real axis

Represents the real part a of the complex number a + bi.

Vertical axis

Imaginary axis

Represents the imaginary part b of the complex number a + bi.

Plotting Examples

3 + 4i

point (3, 4)

−2 + i

point (−2, 1)

5

point (5, 0) — real axis

Why this matters

Treating complex numbers as points (or vectors) in 2D space makes geometric operations — rotation, scaling, finding distance — concrete and visual. The distance from the origin to (a, b) is the modulus r = √(a² + b²), and the angle is the argument θ.

Rectangular Form vs Polar Form

Rectangular FormPolar Form
Notationa + bir(cos θ + i sin θ) or r·cis θ
Componentsreal part a, imaginary part bmodulus r, argument θ
Best forAddition, subtraction, plottingMultiplication, division, powers, roots
Example1 + √3 i2(cos 60° + i sin 60°) = 2·cis 60°

Note: "cis θ" is shorthand for cos θ + i sin θ. Some textbooks use re^(iθ) (Euler form); all three are equivalent.

Converting Rectangular to Polar

Modulus

r = √(a² + b²)

Distance from origin to (a, b)

Argument (base formula)

θ = arctan(b/a)

Then adjust for quadrant (see table)

Quadrant Adjustment Table

QuadrantSigns (a, b)Adjustment
Ia > 0, b > 0None
IIa < 0, b > 0Add 180° (π)
IIIa < 0, b < 0Add 180° (π)
IVa > 0, b < 0Add 360° (2π)

Always verify: r·cos θ should equal a, and r·sin θ should equal b.

Converting Polar to Rectangular

Real part

a = r · cos θ

Imaginary part

b = r · sin θ

Example: Convert 4·cis 150° to rectangular

a = 4 · cos 150° = 4 · (−√3/2) = −2√3

b = 4 · sin 150° = 4 · (1/2) = 2

4·cis 150° = −2√3 + 2i

Worked Conversion Examples

Q II

Example 1: Convert −1 + √3 i to polar form

a = −1, b = √3 (Quadrant II: a < 0, b > 0)

Step 1 — Modulus: r = √((−1)² + (√3)²) = √(1 + 3) = √4 = 2

Step 2 — Base angle: arctan(√3 / −1) = arctan(−√3)

Calculator gives −60°. But a < 0, so add 180°: θ = −60° + 180° = 120°

Step 3 — Verify: 2·cos 120° = 2·(−1/2) = −1 ✓ | 2·sin 120° = 2·(√3/2) = √3 ✓

−1 + √3 i = 2·cis 120° = 2(cos 120° + i sin 120°)

Q III

Example 2: Convert −√2 − √2 i to polar form

a = −√2, b = −√2 (Quadrant III: a < 0, b < 0)

Step 1 — Modulus: r = √((−√2)² + (−√2)²) = √(2 + 2) = √4 = 2

Step 2 — Base angle: arctan(−√2 / −√2) = arctan(1) = 45°

But a < 0, so add 180°: θ = 45° + 180° = 225°

Step 3 — Verify: 2·cos 225° = 2·(−√2/2) = −√2 ✓ | 2·sin 225° = 2·(−√2/2) = −√2 ✓

−√2 − √2 i = 2·cis 225° = 2·cis(5π/4)

Q IV

Example 3: Convert 2 − 2i to polar form

a = 2, b = −2 (Quadrant IV: a > 0, b < 0)

Step 1 — Modulus: r = √(2² + (−2)²) = √(4 + 4) = √8 = 2√2

Step 2 — Base angle: arctan(−2 / 2) = arctan(−1) = −45°

a > 0 and b < 0, so add 360°: θ = −45° + 360° = 315°

Step 3 — Verify: 2√2·cos 315° = 2√2·(√2/2) = 2 ✓ | 2√2·sin 315° = 2√2·(−√2/2) = −2 ✓

2 − 2i = 2√2·cis 315° = 2√2·cis(7π/4)

Multiplication & Division in Polar Form

Multiplication

r₁·cis θ₁ × r₂·cis θ₂

= r₁r₂ · cis(θ₁ + θ₂)

Multiply moduli · add arguments

Division

r₁·cis θ₁ ÷ r₂·cis θ₂

= (r₁/r₂) · cis(θ₁ − θ₂)

Divide moduli · subtract arguments

Worked Example — Multiplication

Compute: 3·cis 40° × 2·cis 110°

= (3 × 2) · cis(40° + 110°)

= 6 · cis 150°

Rectangular check: 6·cos 150° + 6i·sin 150° = −3√3 + 3i

Worked Example — Division

Compute: 10·cis 200° ÷ 5·cis 80°

= (10/5) · cis(200° − 80°)

= 2 · cis 120°

Rectangular: 2(cos 120° + i sin 120°) = −1 + √3 i

De Moivre's Theorem

De Moivre's Theorem

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

(r · cis θ)ⁿ = rⁿ · cis(nθ)

Raise the modulus to the nth power · multiply the argument by n

De Moivre's Theorem is the engine behind computing powers of complex numbers efficiently. Without it, computing (1+i)⁸ would require 7 multiplications with FOIL. With it: convert once, raise r to 8, multiply θ by 8, convert back.

Worked Example: Compute (1 + i)⁸

Step 1 — Convert 1 + i to polar form

r = √(1² + 1²) = √2

θ = arctan(1/1) = 45° (Quadrant I, no adjustment)

1 + i = √2 · cis 45°

Step 2 — Apply De Moivre's with n = 8

(√2 · cis 45°)⁸ = (√2)⁸ · cis(8 × 45°)

= (2^(1/2))⁸ · cis 360°

= 2⁴ · cis 360°

= 16 · cis 360°

Step 3 — Convert back to rectangular

cis 360° = cos 360° + i sin 360° = 1 + 0i = 1

(1 + i)⁸ = 16

nth Roots of Complex Numbers

Formula — All n distinct nth roots

zₖ = r^(1/n) · [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]

for k = 0, 1, 2, ..., n−1

• Every non-zero complex number has exactly n distinct nth roots.

• All n roots lie on a circle of radius r^(1/n).

• The roots are equally spaced by angles of 2π/n (or 360°/n).

Worked Example: Find all 3 cube roots of 8

Step 1 — Write 8 in polar form

8 = 8 + 0i → r = 8, θ = 0°

8 = 8 · cis 0°

Step 2 — Apply the nth root formula with n = 3

zₖ = 8^(1/3) · cis((0° + 360°k)/3) = 2 · cis(120°k)

Step 3 — Compute for k = 0, 1, 2

k = 0: z₀ = 2 · cis 0° = 2

k = 1: z₁ = 2 · cis 120°

= 2(cos 120° + i sin 120°) = 2(−1/2 + i·√3/2)

= −1 + √3 i

k = 2: z₂ = 2 · cis 240°

= 2(cos 240° + i sin 240°) = 2(−1/2 − i·√3/2)

= −1 − √3 i

Verify: All 3 roots cube to 8. The roots form an equilateral triangle on a circle of radius 2, spaced 120° apart.

Exam Tip Cards

Tip 1 — Quadrant Check

After computing θ = arctan(b/a), always verify by plugging back in: r·cos θ must equal a and r·sin θ must equal b. This catches quadrant errors before they cost you points.

Tip 2 — nth Roots Count

The nth root formula always produces exactly n roots (k = 0 through n−1). If you're asked for "all square roots", expect 2 answers. "All cube roots": 3 answers. "All 4th roots": 4 answers. Missing roots means you stopped k too early.

Tip 3 — De Moivre's Shortcut

For large powers (n ≥ 3), always use De Moivre's — never FOIL repeatedly. The conversion: find r and θ, raise r to n, multiply θ by n. A 5-step problem becomes 3 lines.

Frequently Asked Questions

What is the modulus and argument of a complex number?

For a complex number z = a + bi, the modulus (also called absolute value) is r = √(a² + b²) — the distance from the origin to the point (a, b) on the complex plane. The argument θ is the angle that the line segment from the origin to (a, b) makes with the positive real axis, measured counterclockwise. To find it, use θ = arctan(b/a) and then adjust for the correct quadrant: if a < 0 add 180° (or π); if a > 0 and b < 0 add 360° (or 2π). Always verify by checking that r·cos θ = a and r·sin θ = b.

How do you multiply and divide complex numbers in polar form?

Polar form makes multiplication and division straightforward. To multiply z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂): multiply the moduli and add the arguments — z₁z₂ = r₁r₂(cos(θ₁+θ₂) + i sin(θ₁+θ₂)). To divide: divide the moduli and subtract the arguments — z₁/z₂ = (r₁/r₂)(cos(θ₁−θ₂) + i sin(θ₁−θ₂)). This is far more efficient than expanding in rectangular form for powers and roots.

What are the nth roots of a complex number?

Every non-zero complex number z = r(cos θ + i sin θ) has exactly n distinct nth roots. They are given by the formula: zₖ = r^(1/n) · (cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)) for k = 0, 1, 2, ..., n−1. The n roots are equally spaced around a circle of radius r^(1/n) in the complex plane, separated by angles of 2π/n. For example, the 3 cube roots of 8 are 2, 2(cos 120° + i sin 120°) = −1 + i√3, and 2(cos 240° + i sin 240°) = −1 − i√3.

Related Topics

Practice Complex Plane Problems

Interactive problems on polar conversion, De Moivre's Theorem, and nth roots — with step-by-step solutions and private tutoring. Free to try.

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