Polar (modulus-argument) form
A complex number z = x + iy is a point (x, y) on the Argand diagram (a graph where across = real part, up = imaginary part).
- Modulus r = |z| = √(x² + y²). It is the distance from 0.
- Argument θ = arg z. It is the angle from the positive real axis. The principal argument is in (−π, π].
Then x = r cos θ and y = r sin θ, so z = r(cos θ + i sin θ). People often shorten this to r cis θ or r∠θ.
Take care with the quadrant: for z = −1 − i, tan θ = 1, but the point is in the third quadrant, so θ = −3π/4, not π/4.
Exponential (Euler) form
Euler's formula says e^{iθ} = cos θ + i sin θ. So any complex number is z = re^{iθ}.
This form makes rules easy, because powers of e already add: e^{iα}·e^{iβ} = e^{i(α+β)}.
- Put θ = π: e^{iπ} = −1, so e^{iπ} + 1 = 0.
- The conjugate of re^{iθ} is re^{−iθ} (mirror in the real axis).
- cos θ = (e^{iθ} + e^{−iθ})/2 and sin θ = (e^{iθ} − e^{−iθ})/(2i).
Multiplying and dividing in polar form
If z₁ = r₁e^{iα} and z₂ = r₂e^{iβ}:
- z₁z₂ = r₁r₂ e^{i(α+β)} — multiply lengths, add angles.
- z₁/z₂ = (r₁/r₂) e^{i(α−β)} — divide lengths, subtract angles.
So multiplying by i (length 1, angle 90°) just turns a point by 90° anticlockwise. The 3D step 3 shows this as an arrow that stretches and turns.
De Moivre's theorem
[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ) for every integer n (positive, zero or negative).
Why? Multiplying z by itself n times multiplies r n times and adds θ n times. It can be proved for positive n by induction, and for negative n using 1/z = r⁻¹e^{−iθ}.
Use 1: fast powers
(1 + i)¹⁰: r = √2, θ = π/4. So (√2)¹⁰ e^{i·10π/4} = 32 e^{iπ/2} = 32i.
Use 2: multiple-angle formulas
Expand (cos θ + i sin θ)³ with the binomial theorem and compare with cos 3θ + i sin 3θ. Real parts: cos 3θ = cos³θ − 3cos θ sin²θ = 4cos³θ − 3cos θ. Imaginary parts: sin 3θ = 3sin θ − 4sin³θ.
Use 3: powers of cos θ
With z = e^{iθ}: z + 1/z = 2cos θ and zⁿ + 1/zⁿ = 2cos nθ. So (2cos θ)³ = (z + 1/z)³ = z³ + 1/z³ + 3(z + 1/z), giving cos³θ = (cos 3θ + 3cos θ)/4. This helps in integration.
nth roots and roots of unity
To solve zⁿ = w where w = Re^{iφ}: the length must be R^{1/n} and nθ must equal φ plus any whole turn.
z = R^{1/n} e^{i(φ + 2πk)/n}, k = 0, 1, …, n − 1. There are exactly n different roots.
- Roots of unity (w = 1): ω_k = e^{2πik/n}. They sit on the unit circle, 360°/n apart, at the corners of a regular n-gon.
- With ω = e^{2πi/n}, the roots are 1, ω, ω², …, ωⁿ⁻¹, and 1 + ω + … + ωⁿ⁻¹ = 0 (the arrows balance).
- Cube roots of unity: 1, ω = −½ + (√3/2)i, ω² = −½ − (√3/2)i, with ω³ = 1 and 1 + ω + ω² = 0.
- Roots of any w are one root times each root of unity, so they also form a regular polygon, of radius R^{1/n}.
Complex roots of polynomials
Fundamental theorem of algebra: a polynomial of degree n has exactly n complex roots (counting repeats).
- If the coefficients are real, complex roots come in conjugate pairs: if 2 + 3i is a root, so is 2 − 3i.
- Sum of squares now factorises: x² + 4 = (x + 2i)(x − 2i), and x² + y² = (x + iy)(x − iy).
- A conjugate pair a ± bi gives the real quadratic factor x² − 2ax + (a² + b²).
- zⁿ − 1 = (z − 1)(z − ω)…(z − ωⁿ⁻¹), so z³ − 1 = (z − 1)(z² + z + 1).
Try it
In the free-play step, set n = 6 and choose n roots. Before you look, predict: what shape, and what angle between neighbours? Then set θ = 60°, r = 1 and choose Power zⁿ: where does z⁶ land? (Answer: a hexagon, 60°, and z⁶ = 1.) At home: draw a circle, mark 5 points 72° apart with a protractor — you have drawn the fifth roots of unity.
Key formulas and definitions
- z = r(cos θ + i sin θ) = re^{iθ}, r = √(x² + y²), tan θ = y/x (check quadrant)
- z₁z₂ = r₁r₂ e^{i(θ₁+θ₂)}, z₁/z₂ = (r₁/r₂) e^{i(θ₁−θ₂)}
- De Moivre: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
- zⁿ + z⁻ⁿ = 2cos nθ, zⁿ − z⁻ⁿ = 2i sin nθ (z = e^{iθ})
- nth roots of Re^{iφ}: R^{1/n} e^{i(φ+2πk)/n}, k = 0 … n−1
- Roots of unity: 1 + ω + ω² + … + ωⁿ⁻¹ = 0
Worked examples
1. Write z = 1 + √3 i in polar and exponential form.
r = √(1 + 3) = 2. tan θ = √3, first quadrant, so θ = π/3. z = 2(cos π/3 + i sin π/3) = 2e^{iπ/3}.
2. Find (1 + √3 i)⁶.
From above, z = 2e^{iπ/3}. z⁶ = 2⁶ e^{i·6π/3} = 64 e^{2πi} = 64(cos 2π + i sin 2π) = 64.
3. Find (1 − i)⁻⁴.
r = √2, θ = −π/4. z⁻⁴ = (√2)⁻⁴ e^{i·π} = (1/4)(−1) = −1/4.
4. Use De Moivre to show cos 2θ = cos²θ − sin²θ and sin 2θ = 2 sin θ cos θ.
(c + is)² = c² + 2ics − s² (writing c = cos θ, s = sin θ). By De Moivre this equals cos 2θ + i sin 2θ. Real parts: cos 2θ = c² − s². Imaginary parts: sin 2θ = 2sc.
5. Solve z³ = 1 and draw the roots.
z = e^{2πik/3}, k = 0, 1, 2: z = 1, −½ + (√3/2)i, −½ − (√3/2)i. They are corners of an equilateral triangle on the unit circle, 120° apart.
6. Solve z⁴ = −16.
−16 = 16e^{iπ}. Length 16^{1/4} = 2. Angles (π + 2πk)/4 = π/4, 3π/4, 5π/4, 7π/4. Roots: 2e^{iπ/4} = √2 + √2 i, −√2 + √2 i, −√2 − √2 i, √2 − √2 i (a square of radius 2).
7. Show cos³θ = ¼(cos 3θ + 3cos θ).
Let z = e^{iθ}, so z + 1/z = 2cos θ. (z + 1/z)³ = z³ + 3z + 3/z + 1/z³ = (z³ + 1/z³) + 3(z + 1/z) = 2cos 3θ + 6cos θ. So 8cos³θ = 2cos 3θ + 6cos θ, giving cos³θ = ¼(cos 3θ + 3cos θ).
8. A real cubic has roots 2 and 1 + 2i. Find it (leading coefficient 1).
Real coefficients, so 1 − 2i is also a root. Pair gives x² − 2x + 5. Cubic = (x − 2)(x² − 2x + 5) = x³ − 4x² + 9x − 10.
Common mistakes
- Taking θ = tan⁻¹(y/x) without checking the quadrant. For −1 − i the angle is −3π/4, not π/4.
- Writing (r cis θ)ⁿ = r cis nθ — forgetting to raise r to the power n as well.
- Giving only one root of zⁿ = w. There are always n roots; add 2πk before dividing by n.
- Thinking 1 + ω + ω² = 3 or 1. For any n ≥ 2 the sum of all nth roots of unity is 0.