MTH1020 · Analysis of Change
Complex Numbers: Polar Form & De Moivre
Week 6 of Monash MTH1020 Analysis of Change re-expresses complex numbers in polar and exponential form: z = r cis θ = r e^(iθ), where r = |z| is the modulus and θ = arg(z) the argument. From Euler's formula it develops multiplication and division (moduli multiply/divide, arguments add/subtract), De Moivre's theorem zⁿ = rⁿ cis(nθ) for powers, the n distinct n-th roots equally spaced on a circle, loci in the plane, and the Fundamental Theorem of Algebra. These are examined in the mid-semester tests; confirm the exam scope for complex numbers with your cohort on Moodle.
What this chapter covers
- 01Polar form z = r(cos θ + i sin θ) = r cis θ; modulus r = |z| = √(x²+y²)
- 02Argument θ = arg(z) in radians; principal argument Arg(z) ∈ (−π, π]; use a diagram, θ ≠ simply arctan(y/x)
- 03Euler's / exponential form: e^(iθ) = cos θ + i sin θ = cis θ, so z = r e^(iθ); e^(iπ) = −1
- 04Multiplication z₁z₂ = r₁r₂ cis(θ₁+θ₂) and division z₁/z₂ = (r₁/r₂) cis(θ₁−θ₂): moduli multiply/divide, arguments add/subtract
- 05De Moivre's theorem: z = r cis θ ⇒ zⁿ = rⁿ cis(nθ), extending to negative integers
- 06n-th roots: w = ρ cis φ with ρ = r^(1/n), φ = (Arg(z) + 2πk)/n, k = 0, …, n−1 — n roots equally spaced
- 07Loci: |z − z₀| = ρ (circle), Re(z) = c (vertical line), Arg(z) = α (ray); regions from inequalities
- 08Fundamental Theorem of Algebra: every complex polynomial equation has a root in ℂ (ℂ is algebraically closed)
De Moivre's theorem: computing (1 + i)⁸
- +1Find the modulus. For z = 1 + i, r = |z| = √(1² + 1²) = √2.
- +1Find the argument. The point (1, 1) lies in the first quadrant with equal coordinates, so θ = arg(z) = π/4. Hence 1 + i = √2 cis(π/4).
- +1Apply De Moivre's theorem. (1 + i)⁸ = (√2)⁸ cis(8 × π/4) = (√2)⁸ cis(2π). Now (√2)⁸ = (2^(1/2))⁸ = 2⁴ = 16.
- +1Evaluate the cis term. cis(2π) = cos 2π + i sin 2π = 1 + 0i = 1, so (1 + i)⁸ = 16 × 1 = 16.
Key terms
- Modulus (r = |z|)
- The distance of z from the origin in the Argand plane: r = √(x²+y²). Under multiplication moduli multiply; under powers r becomes rⁿ.
- Argument (arg z)
- The directed angle θ, in radians, anticlockwise from the positive real axis. The principal argument Arg(z) lies in (−π, π]; arg(z) = Arg(z) + 2πn.
- Polar / cis form
- z = r(cos θ + i sin θ) = r cis θ. Multiplication and division become 'multiply/divide moduli, add/subtract arguments'.
- Euler's formula
- e^(iθ) = cos θ + i sin θ = cis θ, giving the exponential form z = r e^(iθ). The identity e^(iπ) = −1 is a famous special case.
- De Moivre's theorem
- If z = r cis θ then zⁿ = rⁿ cis(nθ): raise the modulus to the power n and multiply the argument by n. It powers both integer powers and n-th roots.
- n-th roots of a complex number
- A non-zero z has exactly n distinct n-th roots ρ cis φ with ρ = r^(1/n) and φ = (Arg(z) + 2πk)/n for k = 0, …, n−1 — equally spaced 2π/n apart on a circle of radius r^(1/n).
Complex Numbers: Polar Form & De Moivre FAQ
Why is the argument not just arctan(y/x)?
Because arctan only returns angles in (−π/2, π/2), it cannot distinguish the four quadrants. For z = −1 − i the ratio y/x = 1 gives arctan(1) = π/4, but the point sits in the third quadrant, so the true argument is −3π/4 (or 5π/4). Always plot z on the Argand plane, use tan θ = y/x for the reference angle, then adjust for the correct quadrant. This quadrant check is a frequent MTH1020 exam trap.
How does De Moivre's theorem make powers easy?
In polar form a power only touches the modulus and the argument: zⁿ = rⁿ cis(nθ). So instead of multiplying out (1 + i)⁸ by hand, you write 1 + i = √2 cis(π/4), raise √2 to the eighth power and multiply the angle by eight to get 16 cis(2π) = 16. The same theorem run in reverse — dividing the argument — produces the n-th roots.
How many n-th roots does a complex number have, and where are they?
A non-zero complex number has exactly n distinct n-th roots. They all share the modulus r^(1/n), so they sit on a circle of that radius centred at the origin, and their arguments (Arg(z) + 2πk)/n for k = 0, …, n−1 space them evenly 2π/n apart. Plotting them gives a regular n-gon — for the cube roots of unity, an equilateral triangle.
Can Sia help me with polar form and De Moivre?
Yes. Sia can check your modulus and argument (including the quadrant), walk through a De Moivre power or an n-th-root calculation one step at a time, and sketch where the roots fall on the Argand plane. It explains the method and checks your reasoning on your own practice questions; it does not do graded assessment for you, and Monash academic-integrity rules apply.
Exam move
Get fluent moving between x + yi and r cis θ, because polar form is what makes powers and roots tractable. Drill finding the modulus and — carefully — the argument, always plotting the point first so you fix the quadrant rather than trusting arctan. Rehearse De Moivre's theorem for powers and, in reverse, the n-th-root recipe (ρ = r^(1/n), φ = (Arg(z) + 2πk)/n), then sketch the equally spaced roots as a check. Learn the standard loci (circle, line, ray) so you can read a condition like |z − z₀| = ρ instantly. This material is assessed in the mid-semester tests; confirm whether it is on your cohort's final on Moodle. Ask Sia to set fresh powers and roots and verify each line.
Working through Complex Numbers: Polar Form & De Moivre in MTH1020? Sia is AskSia’s AI Mathematics tutor — ask any MTH1020 Complex Numbers: Polar Form & De Moivre question and get a clear, step-by-step explanation grounded in how MTH1020 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.
