IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 HL only ~9 min read

Conversion between Forms of Complex Numbers

By now you’ve met three different ways to write the same complex number — Cartesian (a + bi), polar (r cis θ), and exponential (re^iθ). Each one is best for a different job: Cartesian for adding and subtracting, polar and exponential for multiplying, dividing, and raising to powers. To make the most of all three, you need to be able to flip between them at will. The good news? There are really only two routes to memorise: Cartesian → Polar/Exponential, and the reverse. Polar and exponential are interchangeable — one is just the compact version of the other. Master those routes and every conversion question becomes a clean recipe.

📘 What you need to know

The three forms of the same complex number: z = x + yi = r(cos θ + i sin θ) = re^iθ. The relationship is in the formula booklet.
To convert Cartesian → Polar/Exponential: compute r = √(x² + y²) and the argument θ via the quadrant + reference angle.
To convert Polar/Exponential → Cartesian: evaluate cos θ and sin θ, then expand r(cos θ + i sin θ) into x + yi. So x = r cos θ and y = r sin θ.
Polar ↔ Exponential is essentially copy-paste: r(cos θ + i sin θ) and re^iθ are the same — just take the modulus and argument across.
To go from Exponential → Cartesian directly is not standard — pass through polar form first.
A GDC may have “convert to polar” and “convert to rectangular” functions — useful for Paper 2 sanity checks.
Common exact values to know: cos(π/6) = √3/2, sin(π/6) = 1/2, cos(π/4) = sin(π/4) = √2/2, cos(π/3) = 1/2, sin(π/3) = √3/2.

The three forms — same number, different outfits

Before we talk about converting, let’s lay out the three forms side by side. They all describe the same complex number, but each uses a different “vocabulary”:

Cartesian

x + yi

best for adding & subtracting

Polar (mod-arg)

r(cos θ + i sin θ)

best for multiplication & division

Exponential

re^iθ

best for powers & roots

Polar and exponential always carry the same r and the same θ — only the wrapping changes. Cartesian uses x and y, which are completely different numbers. So most “real” conversion work happens between Cartesian and one of the other two.

Conversion routes between the three forms

Notice the diagram: there’s no direct shortcut from exponential to Cartesian (or back). You always pass through polar form. That’s because Cartesian uses cos θ and sin θ explicitly, while exponential hides them inside e^iθ. The conversion is the same as polar’s, just dressed differently.

Cartesian → Polar (or Exponential)

This is the most common direction in exams. Given z = x + yi, you need r and θ. Use Pythagoras for r and the quadrant rules from the modulus & argument note for θ.

Cartesian → Polar/Exponential r = √(x² + y²) and θ = arg(z)
then z = r(cos θ + i sin θ) = re^iθ

🧭 Recipe — Cartesian to Polar/Exponential

Sketch z on an Argand diagram and identify the quadrant.
Compute r = √(x² + y²) using Pythagoras.
Compute the reference angle α = arctan(|y|/|x|).
Adjust to get the actual argument θ based on the quadrant: Q1 → α, Q2 → π − α, Q3 → −(π − α), Q4 → −α.
Write the answer as r(cos θ + i sin θ) or re^iθ as required.

Polar (or Exponential) → Cartesian

The reverse direction. Given r and θ, find x and y by evaluating the trig functions.

Polar/Exponential → Cartesian x = r cos θ and y = r sin θ
then z = x + yi

🧭 Recipe — Polar/Exponential to Cartesian

If you’re starting from exponential form, first rewrite as polar: re^iθ = r(cos θ + i sin θ).
Evaluate cos θ and sin θ — use exact values where possible (π/6, π/4, π/3, π/2, etc.).
Multiply each by r to get x = r cos θ and y = r sin θ.
Combine into x + yi.

Memorise these exact values: cos(π/6) = √3/2, sin(π/6) = 1/2 · cos(π/4) = sin(π/4) = √2/2 · cos(π/3) = 1/2, sin(π/3) = √3/2. And of course cos(0) = 1, sin(0) = 0, cos(π/2) = 0, sin(π/2) = 1.

Polar ↔ Exponential — the easy swap

Polar and exponential forms carry exactly the same information — modulus r and argument θ. So converting between them is just notation:

Polar ↔ Exponential r(cos θ + i sin θ) ⟺ re^iθ

If you’ve got 4(cos(π/3) + i sin(π/3)), the exponential version is simply 4e^iπ/3. Take the same r, the same θ, and write them in the new wrapper. No calculation needed.

🤔 Why is this swap free?

Because of Euler’s formula: e^iθ = cos θ + i sin θ. So multiplying both sides by r gives re^iθ = r(cos θ + i sin θ). They’re literally equal, just typeset differently. Whatever you can do in one form, you can do in the other.

Worked examples

WE 1

Cartesian to exponential form (Q1)

Express z = 1 + i in exponential form, with the argument in the range −π < θ ≤ π.

Step 1: Sketch — z in Q1 argument will be positive acute Step 2: Modulus r = √(1² + 1²) = √2 Step 3: Argument arctan(1/1) = π/4 Q1 → arg(z) = π/4 Step 4: Write in exponential form z = √2 · e^(iπ/4) in polar form this would be √2(cos(π/4) + i sin(π/4)) — same r and θ, different wrapper

WE 2

Cartesian to polar form (Q3 — careful with sign)

Express z = −2 − 2√3 i in polar form r cis θ, with the argument in the range −π < θ ≤ π.

Step 1: Sketch — both parts negative → Q3 argument will be negative obtuse, between −π and −π/2 Step 2: Modulus r = √((−2)² + (−2√3)²) = √(4 + 12) = √16 = 4 Step 3: Reference angle α = arctan(2√3 / 2) = arctan(√3) = π/3 Step 4: Q3 → θ = −(π − α) θ = −(π − π/3) = −2π/3 z = 4 cis(−2π/3) Q3 always gives a negative argument in the standard range — sketch first, never trust raw arctan

WE 3

Polar to Cartesian form

Express z = 6(cos(π/6) + i sin(π/6)) in Cartesian form a + bi.

Step 1: Identify r and θ r = 6, θ = π/6 Step 2: Look up exact values cos(π/6) = √3/2, sin(π/6) = 1/2 Step 3: Compute x and y x = 6 · √3/2 = 3√3 y = 6 · 1/2 = 3 Step 4: Combine z = 3√3 + 3i always work with exact values where possible — IB markschemes prefer surds over decimals

WE 4

Exponential to Cartesian — go via polar

Express z = 5 e^i·3π/4 in Cartesian form.

Step 1: Convert to polar (Euler’s formula) z = 5(cos(3π/4) + i sin(3π/4)) Step 2: Look up exact values (3π/4 is in Q2) cos(3π/4) = −√2/2, sin(3π/4) = √2/2 Step 3: Compute components x = 5 · (−√2/2) = −5√2/2 y = 5 · (√2/2) = 5√2/2 Step 4: Combine z = −5√2/2 + (5√2/2)i exponential to Cartesian is never direct — always rewrite via polar form first

WE 5

Polar to Cartesian with a negative argument

Express z = 8(cos(−π/3) + i sin(−π/3)) in Cartesian form.

Step 1: Use even/odd rules cos(−π/3) = cos(π/3) = 1/2 (cos is even) sin(−π/3) = −sin(π/3) = −√3/2 (sin is odd) Step 2: Compute x and y x = 8 · 1/2 = 4 y = 8 · (−√3/2) = −4√3 Step 3: Combine z = 4 − 4√3 i a negative argument lands in Q4 (positive real, negative imag) — the answer here confirms it

WE 6

Convert and apply

Two complex numbers are z₁ = 2 + 2i and z₂ = 4 e^i·5π/12.
(a) Write z₁ in exponential form. (b) Find z₁z₂ in exponential form.

Part (a): Convert z₁ to exponential form |z₁| = √(2² + 2²) = √8 = 2√2 arg(z₁) = arctan(2/2) = π/4 (Q1) (a) z₁ = 2√2 · e^(iπ/4) Part (b): Multiply z₁z₂ in exponential form moduli: 2√2 · 4 = 8√2 arguments: π/4 + 5π/12 = 3π/12 + 5π/12 = 8π/12 = 2π/3 (b) z₁z₂ = 8√2 · e^(i·2π/3) converting first lets you use the index-law shortcut for multiplication — much faster than expanding (2+2i)·(Cartesian form of z₂)

💡 Top tips

Use the right form for the right job. Cartesian for adding/subtracting, polar/exponential for multiplying, dividing, and powers.
Always sketch when computing arguments from Cartesian. The sketch picks the quadrant; the quadrant decides the sign of θ.
Polar ↔ exponential is free. Just copy r and θ across. No arithmetic.
Exponential → Cartesian goes through polar. First rewrite re^iθ as r(cos θ + i sin θ), then expand.
Memorise common exact values for cos and sin at π/6, π/4, π/3, π/2 (and their multiples). They cut working time in half.
Use cos(−θ) = cos θ and sin(−θ) = −sin θ when the argument is negative — saves a sketch.
Your GDC may have polar/rectangular conversion buttons. Useful on Paper 2 to check answers — but on Paper 1 you’ll need to show working.

⚠ Common mistakes

Trusting raw arctan(y/x) on the calculator. The calculator can’t tell Q2 from Q4 — always sketch first.
Skipping the polar step when going from exponential to Cartesian. There’s no direct shortcut; you must use Euler’s formula in between.
Using decimal approximations when exact values exist. If θ = π/3, write cos(π/3) = 1/2, not 0.5 — IB markschemes prefer exact form.
Forgetting that r must be positive. If you ever derive a negative r, fold the minus into the angle by adding π.
Confusing (r, θ) for (x, y). The point with Cartesian (3, 4) is not the same as the point with polar (r, θ) = (3, 4). Different coordinate systems.
Using degrees instead of radians. All HL complex-number arguments are in radians. Check your calculator mode.
Forgetting to check the argument range. If a question specifies −π < θ ≤ π or 0 ≤ θ < 2π, make sure your final answer falls inside.

Once you can move comfortably between all three forms, the rest of the Further Complex Numbers section becomes much smoother. The next big payoff is De Moivre’s theorem, which uses exponential and polar form to take huge powers of complex numbers in a single line. Before that, the next note covers complex roots of polynomials — the area where conjugate pairs and quadratic factoring really earn their keep.

Need help with Conversion between Forms?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →