8.5Complex Numbers in Polar Form
Write z = a + bi in polar form r(cosθ + i sinθ). Multiply and divide complex numbers in polar form. Apply De Moivre's Theorem zⁿ = rⁿ(cos nθ + i sin nθ) to find powers and roots.
Polar form makes multiplication and powers of complex numbers geometric — you multiply moduli and add arguments. De Moivre's Theorem is the key to finding nth roots of complex numbers and appears in electrical engineering.
Essential Question
How does writing a complex number in polar form r(cosθ + i sinθ) make multiplication, division, and finding powers and roots dramatically simpler than working in rectangular form?
Lesson Overview
The complex number z = a + bi plotted in the complex plane. r is the modulus (distance from origin), θ is the argument.
Every complex number z = a + bi can be written in polar form: z = r(cosθ + i sinθ), also written r cis θ.
- Modulus: r = |z| = √(a² + b²) — the distance from the origin to z in the complex plane.
- Argument: θ = arctan(b/a), adjusted for the correct quadrant. The principal argument satisfies θ ∈ (−π, π].
- 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.
- De Moivre's Theorem: [r(cosθ + i sinθ)]ⁿ = rⁿ(cos nθ + i sin nθ)
- nth roots: The n distinct roots of z = r cis θ are r^(1/n) cis((θ + 2πk)/n) for k = 0, 1, …, n − 1. The roots are equally spaced around a circle of radius r^(1/n).
Worked Examples
Convert z = −3 + 3i to polar form.
Step 1: Find r. r = √((−3)² + 3²) = √(9 + 9) = √18 = 3√2
Step 2: Find θ. The point (−3, 3) is in Quadrant II.
arctan(3 / −3) = arctan(−1) = −π/4, but we adjust for Q II:
θ = π − π/4 = 3π/4
Step 3: Write polar form.
Convert z = 4(cos(5π/6) + i sin(5π/6)) to rectangular form.
Step 1: Evaluate cos(5π/6) = −√3/2 and sin(5π/6) = 1/2.
Step 2: a = r cosθ = 4 · (−√3/2) = −2√3
b = r sinθ = 4 · (1/2) = 2
Multiply z₁ = 2(cos(π/4) + i sin(π/4)) and z₂ = 3(cos(π/3) + i sin(π/3)).
Multiply moduli: r = 2 · 3 = 6
Add arguments: θ = π/4 + π/3 = 3π/12 + 4π/12 = 7π/12
Use De Moivre's Theorem to find (1 + i)⁸.
Step 1: Convert 1 + i to polar form.
r = √(1² + 1²) = √2, θ = arctan(1/1) = π/4
So 1 + i = √2 cis(π/4)
Step 2: Apply De Moivre: (√2)⁸ cis(8 · π/4)
(√2)⁸ = (2^(1/2))⁸ = 2⁴ = 16
8 · π/4 = 2π → cis(2π) = cos(2π) + i sin(2π) = 1 + 0i
Find all cube roots of z = 8(cos(π/2) + i sin(π/2)).
r = 8, θ = π/2, n = 3.
Root modulus: 8^(1/3) = 2.
Root arguments: (π/2 + 2πk)/3 for k = 0, 1, 2.
k=0: (π/2)/3 = π/6
k=1: (π/2 + 2π)/3 = (5π/2)/3 = 5π/6
k=2: (π/2 + 4π)/3 = (9π/2)/3 = 3π/2
Guided Practice
Convert z = −2 − 2√3 i to polar form.
Hint: Find r = √(4 + 12) = 4. The point (−2, −2√3) is in Quadrant III — arctan(−2√3 / −2) = arctan(√3) = π/3, so θ = π + π/3 = 4π/3.
Convert 5(cos(7π/6) + i sin(7π/6)) to rectangular form.
Hint: cos(7π/6) = −√3/2 and sin(7π/6) = −1/2. Multiply each by 5.
Divide z₁ = 6(cos(3π/4) + i sin(3π/4)) by z₂ = 2(cos(π/4) + i sin(π/4)).
Hint: Divide moduli: 6/2 = 3. Subtract arguments: 3π/4 − π/4 = π/2.
Use De Moivre's Theorem to compute (√3 + i)⁶.
Hint: Convert to polar: r = √(3 + 1) = 2, θ = arctan(1/√3) = π/6. Apply De Moivre: 2⁶ cis(6 · π/6) = 64 cis(π).
Find all fourth roots of 16.
Hint: Write 16 = 16 cis(0). Root modulus: 16^(1/4) = 2. Arguments: (0 + 2πk)/4 = πk/2 for k = 0, 1, 2, 3.
Key Vocabulary
Polar form
A way of writing a complex number as z = r(cosθ + i sinθ) = r cis θ, using its distance from the origin (r) and angle (θ).
Example: z = 2(cos(π/3) + i sin(π/3))
Modulus
The modulus of z = a + bi is r = |z| = √(a² + b²). It is the distance from the origin to z in the complex plane.
Example: |3 + 4i| = √(9 + 16) = 5
Argument
The angle θ that the line from the origin to z makes with the positive real axis. Computed as θ = arctan(b/a), adjusted for quadrant.
Example: arg(−1 + i) = 3π/4
Principal argument
The unique argument θ satisfying −π < θ ≤ π. Denoted Arg(z).
Example: Arg(−1) = π
cis notation
Shorthand: cis θ = cosθ + i sinθ. So r cis θ = r(cosθ + i sinθ).
Example: 3 cis(π/2) = 3i
De Moivre's Theorem
[r(cosθ + i sinθ)]ⁿ = rⁿ(cos nθ + i sin nθ). Used to compute powers and roots of complex numbers.
Example: (cos(π/4) + i sin(π/4))⁴ = cos(π) + i sin(π) = −1
nth roots of a complex number
The n distinct solutions to wⁿ = z. Given z = r cis θ, the roots are r^(1/n) cis((θ + 2πk)/n) for k = 0, 1, …, n−1.
Example: Cube roots of 8: 2 cis(0), 2 cis(2π/3), 2 cis(4π/3)
Complex plane
A coordinate plane where the horizontal axis represents the real part and the vertical axis represents the imaginary part of a complex number.
Example: z = 3 + 4i is plotted at the point (3, 4)
Practice Quiz
Interactive Practice — 5 Questions
What is the modulus of z = −3 + 4i?
z = 2(cos(π/6) + i sin(π/6)) in rectangular form is:
When multiplying two complex numbers in polar form, you:
De Moivre's Theorem: [2(cos(π/4) + i sin(π/4))]⁴ = ?
The number of distinct nth roots of any nonzero complex number is:
Common Mistakes
Using θ = arctan(b/a) without adjusting for the quadrant of z.
arctan only returns values in (−π/2, π/2). Always check which quadrant (a, b) lies in and adjust θ accordingly.
In De Moivre's Theorem, raising only the angle to the nth power: [r cis θ]ⁿ = r cis(nθ).
Both r AND θ are affected: [r cis θ]ⁿ = rⁿ cis(nθ). Don't forget to raise r to the nth power.
When finding nth roots, computing only k = 0 and stopping.
There are n distinct roots. Use k = 0, 1, 2, …, n−1 to find all of them.
Confusing the multiplication rule with the division rule.
Multiplication: multiply r, ADD θ. Division: divide r, SUBTRACT θ. Keep them straight.
Math Tips
Polar form makes multiplication and division elegant: multiply/divide the moduli, add/subtract the arguments.
De Moivre's Theorem is the key to computing large powers of complex numbers — convert to polar first, apply the theorem, then convert back.
The n nth-roots of a complex number are equally spaced around a circle of radius r^(1/n), separated by 2π/n radians.
The principal argument is in (−π, π]. When the problem says "find the argument," give the principal value unless asked for all representations.
Quick check: |z₁ · z₂| = |z₁| · |z₂|. If your product has a different modulus than r₁ · r₂, you made an error.