Practice Maths

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

Hot Tip: When finding arg(z) for z = a + bi, always identify the quadrant first. arctan(b/a) gives an angle in (−π/2, π/2), which is only correct for the first and fourth quadrants. For second/third quadrant, add or subtract π accordingly.

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

  1. 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

  2. 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)

  3. 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;

  4. 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

  5. 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.

  6. Understanding

    Q6 — Square Roots in Polar Form

    Find the two square roots of z = 4 cis(π/3), leaving answers in polar form.

  7. 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)

  8. Understanding

    Q8 — Modulus and Argument of a Product

    Given z = 1 − i and w = √3 + i, find |zw| and arg(zw) without expanding zw.

  9. 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.

  10. 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.