Polar Form and De Moivre’s Theorem
Key Terms
- Polar form
- z = r cis θ where r = |z| and θ = arg(z); cis θ = cos θ + i sin θ.
- Multiplication in polar form
- r1 cis θ1 × r2 cis θ2 = r1r2 cis(θ1 + θ2): multiply moduli, ADD arguments.
- Division in polar form
- (r1 cis θ1) / (r2 cis θ2) = (r1/r2) cis(θ1 − θ2): divide moduli, SUBTRACT arguments.
- De Moivre’s theorem
- (r cis θ)n = rn cis(nθ) for any integer n.
- Principal argument
- θ ∈ (−π, π]; always identify the correct quadrant using the signs of a and b.
- Converting to polar
- r = √(a²+b²); θ = arctan(b/a) adjusted to the correct quadrant.
Polar Form & De Moivre’s Theorem — Key Facts
Polar (modulus-argument) form: z = r(cosθ + i sinθ) = r cisθ
Modulus: r = |z| = √(a² + b²) Argument: θ = arg(z), the angle from the positive real axis
Converting to polar: given z = a + bi, find r = √(a²+b²) then θ = arctan(b/a) adjusted for the correct quadrant
Multiplication: r&sub1;cisθ&sub1; × r&sub2;cisθ&sub2; = r&sub1;r&sub2; cis(θ&sub1; + θ&sub2;)
Division: r&sub1;cisθ&sub1; ÷ r&sub2;cisθ&sub2; = (r&sub1;/r&sub2;) cis(θ&sub1; − θ&sub2;)
De Moivre’s Theorem: [r cisθ]n = rn cis(nθ) for any integer n
Principal argument: θ ∈ (−π, π] is conventionally used
Worked Example 1 — Converting to Polar Form
Write z = 1 + √3 i in polar form r cisθ.
Step 1 — Find r: r = √(1² + (√3)²) = √(1 + 3) = √4 = 2
Step 2 — Find θ: a = 1 > 0, b = √3 > 0 (first quadrant). tanθ = √3/1 = √3, so θ = π/3
Answer: z = 2 cis(π/3)
Worked Example 2 — Applying De Moivre’s Theorem
Find (1 + i)8.
Step 1 — Convert to polar: r = √(1²+1²) = √2, θ = π/4 (first quadrant)
So 1 + i = √2 cis(π/4)
Step 2 — Apply De Moivre: (1 + i)8 = (√2)8 cis(8 × π/4) = (√2)8 cis(2π)
Step 3 — Simplify: (√2)8 = (21/2)8 = 24 = 16. cis(2π) = cos(2π) + i sin(2π) = 1.
Answer: (1 + i)8 = 16
A New Way to Describe Complex Numbers
Every complex number z = a + bi can be visualised as the point (a, b) on the Argand diagram. The Cartesian form tells us the horizontal and vertical coordinates. But there is another equally natural way to locate a point in a plane: give its distance from the origin and the angle it makes with the positive real axis. This is the polar (or modulus-argument) form, and it turns out to be far more powerful for multiplication, division, and finding powers of complex numbers.
Modulus and Argument
The modulus of z = a + bi is r = |z| = √(a² + b²). This is simply the distance from the origin to the point (a, b), calculated by Pythagoras’ theorem. The argument θ = arg(z) is the angle (measured anticlockwise from the positive real axis) to the line segment joining the origin to z. By convention we use the principal argument where θ ∈ (−π, π].
To find the argument: compute the reference angle α = arctan(|b/a|) and then adjust for the quadrant. If z is in Q1: θ = α. Q2: θ = π − α. Q3: θ = −(π − α) = α − π. Q4: θ = −α.
The cis Notation
We write cisθ as shorthand for cosθ + i sinθ. So any complex number with modulus r and argument θ can be written as z = r cisθ. Converting back to Cartesian form: a = r cosθ, b = r sinθ. This connection between polar coordinates and trigonometry is why this form is so useful.
Multiplication and Division in Polar Form
This is where polar form shines. If z&sub1; = r&sub1; cisθ&sub1; and z&sub2; = r&sub2; cisθ&sub2;, then:
z&sub1;z&sub2; = r&sub1;r&sub2; cis(θ&sub1; + θ&sub2;) — multiply the moduli, add the arguments
z&sub1;/z&sub2; = (r&sub1;/r&sub2;) cis(θ&sub1; − θ&sub2;) — divide the moduli, subtract the arguments
This is far simpler than expanding in Cartesian form. Geometrically, multiplication by a complex number simultaneously scales and rotates the original number.
De Moivre’s Theorem
Applying the multiplication rule repeatedly: (r cisθ)2 = r² cis(2θ), (r cisθ)3 = r³ cis(3θ), and in general:
[r cisθ]n = rn cis(nθ) for any integer n.
This theorem is extraordinary: it says that to raise a complex number to the n-th power, you simply raise the modulus to the n-th power and multiply the argument by n. What would be a lengthy algebraic expansion collapses to two arithmetic steps. De Moivre’s theorem also extends to rational exponents, making it the standard tool for finding square roots and cube roots of complex numbers in polar form.
Finding Roots of Complex Numbers
To find the n-th roots of z = r cisθ, use: z1/n = r1/n cis((θ + 2kπ)/n) for k = 0, 1, 2, …, n−1. This gives exactly n distinct roots, equally spaced around a circle of radius r1/n in the Argand plane. For example, the three cube roots of 8 are found by writing 8 = 8 cis(0) and computing: 2 cis(0) = 2, 2 cis(2π/3) = −1 + √3 i, 2 cis(4π/3) = −1 − √3 i.
Why This Matters
De Moivre’s theorem underpins trigonometric identities (you can derive cos(3θ) and sin(3θ) from (cisθ)3), signal processing, electrical engineering (phasors), and quantum mechanics. In this course, it is the gateway to solving polynomial equations completely over the complex numbers.
Mastery Practice
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Fluency
Q1 — Converting to Polar Form
Write each complex number in polar form r cisθ, giving θ as an exact value:
(a) z = √3 + i (b) z = −1 + i (c) z = −2 (d) z = −i
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Fluency
Q2 — Converting from Polar to Cartesian
Write each complex number in Cartesian form a + bi:
(a) z = 4 cis(0) (b) z = 2 cis(π/2) (c) z = 3 cis(2π/3) (d) z = 6 cis(−π/6)
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Fluency
Q3 — Multiplying and Dividing in Polar Form
Given z&sub1; = 3 cis(π/4) and z&sub2; = 2 cis(π/6), find: (a) z&sub1;z&sub2; (b) z&sub1;/z&sub2;
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Fluency
Q4 — Applying De Moivre’s Theorem (Basic)
Use De Moivre’s theorem to evaluate: (a) [2 cis(π/3)]3 (b) [cis(π/6)]12 (c) [√2 cis(π/4)]4
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Understanding
Q5 — Powers of Complex Numbers
Find (1 + i)6 by first converting to polar form, then applying De Moivre’s theorem. Express your answer in Cartesian form.
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Understanding
Q6 — Square Roots in Polar Form
Find the two square roots of z = 4 cis(π/3), leaving answers in polar form.
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Understanding
Q7 — Geometric Interpretation
Explain geometrically what happens to a complex number z when you:
(a) Multiply it by i (b) Multiply it by −1 (c) Multiply it by 2 cis(π/3)
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Understanding
Q8 — Modulus and Argument of a Product
Given z = 1 − i and w = √3 + i, find |zw| and arg(zw) without expanding zw.
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Problem Solving
Q9 — Cube Roots of Unity
Find all three cube roots of 1 (i.e., solve z3 = 1 over ℂ). Express each root in Cartesian form.
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Problem Solving
Q10 — Using De Moivre to Derive a Trigonometric Identity
Using De Moivre’s theorem with n = 2, show that cos(2θ) = cos²θ − sin²θ and sin(2θ) = 2 sinθ cosθ.