IB Maths AI HL Further Complex Numbers Paper 1 & 2 Form conversion ~8 min read

Conversion between Forms of Complex Numbers

A complex number has three faces: Cartesian x + yi, polar r(cos θ + isin θ), and exponential re^iθ. All represent the same number; the polar form is the bridge between the other two.

📘 What you need to know

All three forms are equivalent: z = x + yi = r(cos θ + i sin θ) = re^iθ.
The relationship is in the formula booklet.
Out of Cartesian: compute r = √(x²+y²) and θ = arg z (quadrant-aware).
Into Cartesian: evaluate cos θ and sin θ (often exact), then multiply through by r.
You cannot go directly from exponential to Cartesian — always pass through polar.
Mind the range: −π < θ ≤ π or 0 ≤ θ < 2π, depending on what the question asks.

The three forms and how they relate

The three forms answer the same question in different languages. Cartesian gives the point’s coordinates. Polar gives its modulus and the cosine/sine of its argument. Exponential compresses polar into a single exponential expression. They share the same r and θ; only the wrapper around them changes.

Cartesian ↔ polar uses r, θ and the cos/sin values. Polar ↔ exponential is a pure notation swap. Cartesian ↔ exponential always passes through polar.

The three equivalent forms z = x + yi = r(cos θ + i sin θ) = re^iθ = r cis θ r = √(x² + y²), θ = arg z

Converting out of Cartesian

Starting with z = x + yi, find the modulus r = √(x²+y²) and the argument θ using a sketch and the quadrant rule. Once you have r and θ, both polar and exponential are written directly: z = r(cos θ + i sin θ) = re^iθ.

Watch the range: principal-value questions use −π < θ ≤ π, while exponential-form questions often ask for 0 ≤ θ < 2π. Add or subtract 2π if θ doesn’t fit.

Converting into Cartesian

Going the other way, evaluate cos θ and sin θ — usually exact at standard angles like π/6, π/4, π/3, π/2 — then multiply each by r to get x and y. Exponential form does not convert directly to Cartesian. First write re^iθ as r(cos θ + i sin θ), then expand.

🧭 Recipe — converting between any two forms

Identify the starting form: Cartesian, polar, or exponential.
Identify the target form: Cartesian, polar, or exponential.
If neither end is polar, route through polar — never convert exponential straight to Cartesian.
From Cartesian: compute r and θ. To Cartesian: evaluate cos θ, sin θ, then expand.
Adjust θ into the requested range by adding or subtracting 2π.

Worked examples

WE 1

Cartesian to both polar and exponential

Express z = −1 + √3 i in (a) polar form and (b) exponential form.

r = √((−1)² + (√3)²) = √4 r = 2 sketch: Q2, positive obtuse — α = tan⁻¹(√3) = π/3 θ = π − π/3 = 2π/3 (a) 2(cos 2π/3 + i sin 2π/3) · (b) 2 e^{i 2π/3} same r, same θ — the two forms differ only in wrapper.

WE 2

Polar to Cartesian (quadrant 3)

Write z = 4(cos(7π/6) + i sin(7π/6)) in Cartesian form.

7π/6 lies in quadrant 3 — both cos and sin negative cos 7π/6 = −√3/2, sin 7π/6 = −1/2 multiply through by r = 4 z = 4(−√3/2 − i/2) = −2√3 − 2i z = −2√3 − 2i a Q3 argument gives both parts negative — quick sanity check.

WE 3

Exponential to Cartesian via polar

Express z = √2 e^{i 5π/4} in Cartesian form.

first rewrite in polar form z = √2 (cos 5π/4 + i sin 5π/4) 5π/4 in Q3: cos = −√2/2, sin = −√2/2 z = √2 (−√2/2 − i√2/2) = −1 − i z = −1 − i e^{i θ} can’t be expanded directly — the polar middle step is essential.

WE 4

Cartesian to exponential with 0 ≤ θ < 2π

Write z = −2 − 2√3 i in the form re^iθ with 0 ≤ θ < 2π.

r = √(4 + 12) = √16 = 4 sketch: x < 0, y < 0 ⇒ Q3, negative obtuse principal arg α = tan⁻¹(2√3/2) = π/3; θ_princ = −(π − π/3) = −2π/3 range asked: 0 ≤ θ < 2π, so add 2π θ = −2π/3 + 2π = 4π/3 z = 4 e^{i 4π/3} always check the asked range — 4π/3 is the Q3 angle in [0, 2π).

WE 5

Polar to exponential and to Cartesian

Let z = 6 cis(−π/4). (a) Write z in exponential form. (b) Write z in Cartesian form.

(a) polar → exponential: same r, same θ z = 6 e^{−i π/4} (b) evaluate cos(−π/4) = √2/2, sin(−π/4) = −√2/2 z = 6(√2/2 − i√2/2) = 3√2 − 3√2 i (a) 6 e^{−i π/4} · (b) 3√2 − 3√2 i going polar → exponential is a free conversion — the maths is identical.

WE 6

Full question: convert, multiply, convert back

Let z₁ = √3 − i and z₂ = 4 e^iπ/2. (a) Write z₁ in exponential form with 0 ≤ θ < 2π. (b) Find z₁z₂ in the same form. (c) Write z₁z₂ in Cartesian form.

(a) r₁ = √(3+1) = 2; Q4 ⇒ θ = −π/6 (principal) in [0, 2π): −π/6 + 2π = 11π/6 z₁ = 2 e^{i 11π/6} (b) multiply moduli, add arguments r = 2 · 4 = 8; θ = 11π/6 + π/2 = 14π/6 = 7π/3 7π/3 > 2π, so subtract 2π 7π/3 − 6π/3 = π/3 ⇒ z₁z₂ = 8 e^{i π/3} (c) 8(cos π/3 + i sin π/3) = 8(1/2 + i√3/2) (a) 2 e^{i 11π/6} · (b) 8 e^{i π/3} · (c) 4 + 4√3 i check: (√3 − i)(4i) = 4√3 i − 4i² = 4 + 4√3 i. The Cartesian route confirms it.

💡 Top tips

Polar ↔ exponential is free — same r, same θ, only the wrapper changes.
Sketch before converting from Cartesian — the quadrant pins down θ.
Standard angles (π/6, π/4, π/3, π/2 and multiples) give exact cos and sin values — memorise the table.
Out of Cartesian and back: same procedure on r and θ — the wrapper is interchangeable.
Your GDC may convert between forms (look for ‘polar’ or ‘rectangular’ modes) — useful for checking.

⚠ Common mistakes

Going exponential straight to Cartesian — you must route through polar (cos and sin).
Forgetting to check the asked range for θ — principal vs 0 ≤ θ < 2π.
Using only tan⁻¹ — that gives the reference angle, not the argument; quadrant adjustment is needed.
Approximating cos/sin when an exact value (e.g. √3/2) was available.
Mixing degrees and radians — complex-number arguments are in radians.

Next up: Frequency & Phase of Trig Functions — using exponential form to add two sinusoidal signals with the same frequency. A surprising application of complex numbers to AC circuits.

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