IB Maths AI HL Further Complex Numbers Paper 1 & 2 Polar form ~8 min read

Modulus-Argument (Polar) Form

Polar form rewrites a complex number using its modulus and argument: z = r(cos θ + isin θ). It looks heavier than Cartesian, but multiplying, dividing and conjugating each become a one-line rule on r and θ.

📘 What you need to know

Polar form: z = r(cos θ + i sin θ), with r = |z| and θ = arg z; also written r cis θ.
The form is given in the formula booklet.
Cartesian → polar: compute r and find θ by quadrant.
Polar → Cartesian: evaluate cos θ and sin θ, then multiply through by r.
Products multiply moduli, add arguments; quotients divide moduli, subtract arguments.
The conjugate of r cis θ is r cis(−θ) — same modulus, opposite argument.

Polar form and converting between forms

The same complex number can be written two ways. Cartesian form z = x + yi uses the point’s coordinates; polar form uses its distance r from O and the angle θ it turns from the positive real axis. Both describe exactly the same point on the Argand diagram.

For z = √3 + i: the modulus r = 2 is the length of the arrow, and the argument θ = π/6 is the angle at O. Cartesian and polar describe the same point.

Modulus-argument (polar) form z = r(cos θ + i sin θ) = r cis θ where r = |z| = √(x² + y²) and θ = arg z use the quadrant rule to place θ in the correct range

Multiplying and dividing in polar form

This is where polar form pays off. If z₁ = r₁ cis θ₁ and z₂ = r₂ cis θ₂, the product is r₁r₂ cis(θ₁+θ₂) and the quotient is (r₁/r₂) cis(θ₁−θ₂) — no bracket expansion needed.

The same rule covers powers: zⁿ = rⁿ cis(nθ), since powers are just repeated multiplication.

Range adjustment: products and quotients can push the new argument outside −π < θ ≤ π (or 0 ≤ θ < 2π, depending on the question). Add or subtract 2π until the argument fits. For instance −7π/6 + 2π = 5π/6 lies in (−π, π].

The complex conjugate in polar form

The conjugate flips the sign of the imaginary part, which on the Argand diagram is a reflection in the real axis. Reflecting reverses the angle but keeps the length: z* has the same modulus and the opposite argument.

Show negative arguments inside the brackets, not outside. The conjugate of 2 cis(π/3) is 2(cos(−π/3) + i sin(−π/3)) — never 2(cos(π/3) − i sin(π/3)). Polar form requires a plus between the cos and the i sin.

🧭 Recipe — converting Cartesian to polar form

Find the modulus: r = √(x² + y²).
Sketch z on an Argand diagram and identify its quadrant.
Reference angle: α = tan⁻¹(|y| / |x|).
Adjust for the quadrant to get θ in −π < θ ≤ π.
Write z = r(cos θ + i sin θ) — sign on θ goes inside the cos and sin.

Worked examples

WE 1

Cartesian to polar in quadrant 2

Write z = −1 + i in polar form.

modulus r = √(x² + y²) r = √((−1)² + 1²) = √2 sketch: x < 0, y > 0 ⇒ quadrant 2 (positive obtuse) α = tan⁻¹(1/1) = π/4; θ = π − π/4 = 3π/4 z = √2 (cos 3π/4 + i sin 3π/4) positive and obtuse — exactly what quadrant 2 should give.

WE 2

Polar with negative argument to Cartesian

Write z = 8(cos(−π/4) + i sin(−π/4)) in Cartesian form.

evaluate cos and sin of −π/4 cos(−π/4) = √2 / 2, sin(−π/4) = −√2 / 2 multiply through by r = 8 z = 8(√2/2 − i√2/2) = 4√2 − 4√2 i z = 4√2 − 4√2 i a negative argument lands in quadrant 4 — the answer’s positive real and negative imaginary parts confirm it.

WE 3

Conjugate in polar form

Let z = 3(cos(2π/5) + i sin(2π/5)). Write z* in polar form, and find zz*.

conjugate: same modulus, opposite argument z* = 3(cos(−2π/5) + i sin(−2π/5)) zz* via polar: multiply moduli, add arguments zz* = 3 · 3 · cis(2π/5 + (−2π/5)) = 9 cis(0) z* = 3 cis(−2π/5) · zz* = 9 zz* always lands on the positive real axis — it equals r² = |z|².

WE 4

Product in polar form

Find the product of z₁ = 5 cis(π/3) and z₂ = 2 cis(π/4) in the form r(cos θ + i sin θ).

multiply moduli, add arguments r = 5 × 2 = 10 θ = π/3 + π/4 = 4π/12 + 3π/12 = 7π/12 z₁z₂ = 10(cos 7π/12 + i sin 7π/12) a common denominator of 12 makes the addition clean — no decimals needed.

WE 5

Quotient with range adjustment

Find z₁ / z₂ where z₁ = 10 cis(−2π/3) and z₂ = 5 cis(π/2), giving your answer in the form r cis θ with −π < θ ≤ π.

divide moduli, subtract arguments r = 10 ÷ 5 = 2 θ = −2π/3 − π/2 = −4π/6 − 3π/6 = −7π/6 −7π/6 < −π — add 2π to bring into range −7π/6 + 2π = 5π/6 z₁/z₂ = 2 cis(5π/6) argument too negative this time — add 2π, the same angle restated.

WE 6

Full question: polar form and a fourth power

Let z = 1 − i. (a) Write z in polar form. (b) Hence find z⁴ in polar form with −π < θ ≤ π. (c) Write z⁴ in Cartesian form.

(a) r = √(1 + 1) = √2; quadrant 4 ⇒ θ = −π/4 z = √2 (cos(−π/4) + i sin(−π/4)) (b) z⁴ = (√2)⁴ cis(4 · (−π/4)) = 4 cis(−π) −π sits on the open boundary — add 2π −π + 2π = π ⇒ z⁴ = 4(cos π + i sin π) (c) cos π = −1, sin π = 0 (a) √2 cis(−π/4) · (b) 4(cos π + i sin π) · (c) −4 cross-check: (1 − i)² = −2i, so (1 − i)⁴ = (−2i)² = −4. The two routes agree.

💡 Top tips

Sketch first when converting from Cartesian — the quadrant fixes the sign of θ.
Put the − inside the cos and sin: cos(−θ) + i sin(−θ) — never in front.
For products: multiply moduli, add arguments. For quotients: divide moduli, subtract.
For powers: zⁿ = rⁿ cis(nθ) — saves expanding brackets n times.
Out of range? Add or subtract 2π until θ fits the asked interval.

⚠ Common mistakes

Writing r(cos θ − i sin θ) — polar form needs a +; put the minus on θ inside cos and sin.
Skipping the quadrant adjustment — tan⁻¹ gives only the reference angle.
Multiplying arguments instead of adding (or vice versa for moduli).
Leaving an out-of-range argument — always check it fits the asked range.
Using degrees — polar arguments are in radians.

Next up: Exponential (Euler’s) Form — the polar form’s compact cousin, z = re^iθ. Multiplication, division and powers reduce to ordinary index laws.

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