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

Exponential (Euler’s) Form

If polar form felt slick, exponential form is downright magical. The same content — modulus and argument — gets squashed into the compact expression z = re^iθ. The big surprise (and there’s no avoiding the surprise) is that the imaginary unit shows up in an exponent. Once you accept this, multiplication becomes “multiply moduli, add exponents” — which is just normal index laws. Powers become trivial. Roots become almost trivial. This is the form Euler discovered in the 18th century, and it’s the form you’ll use in nearly every advanced complex-number question. The big idea: the same “scale and rotate” picture, written in the cleanest possible notation.

📘 What you need to know

Exponential (Euler’s) form: z = re^iθ, where r = |z| and θ = arg(z). It’s in the formula booklet.
This is the same complex number as r(cos θ + i sin θ) — just written more compactly.
Multiplication: r₁e^iθ₁ · r₂e^iθ₂ = r₁r₂e^{i(θ₁+θ₂)}. Standard index laws apply.
Division: (r₁e^iθ₁) / (r₂e^iθ₂) = (r₁/r₂)e^{i(θ₁−θ₂)}.
Powers: (re^iθ)ⁿ = rⁿe^inθ — straight from index laws.
Special values to memorise: e^iπ = −1, e^2πi = 1, e^iπ/2 = i.
Euler’s identity: e^iπ + 1 = 0 — often called the most beautiful equation in mathematics.
The position of i in the exponent doesn’t matter: re^iθ, re^θi, re^(θ)i — all the same.

What is exponential form?

Exponential form takes the polar expression r(cos θ + i sin θ) and rewrites it using a single exponential:

Exponential (Euler’s) form z = re^iθ where r = |z| and θ = arg(z) ✓ in the formula booklet

The connection between the two forms comes from Euler’s formula, one of the most surprising results in mathematics:

Euler’s formula — the bridge between the forms e^iθ = cos θ + i sin θ

So multiplying both sides by r gives re^iθ = r(cos θ + i sin θ) — the same complex number, just written two different ways.

🤔 Why is the imaginary unit in the exponent?

The proof goes through Maclaurin series (which you’ll meet in the calculus topic). When you expand e^x, cos θ, and sin θ as infinite polynomials and substitute x = iθ into e^x, the result simplifies exactly to cos θ + i sin θ. So Euler’s formula isn’t a definition — it’s a theorem, and a stunning one. For now, just accept it and use it.

The geometric picture

For any angle θ, the complex number e^iθ sits on the unit circle in the complex plane — at distance 1 from the origin, in direction θ. Multiplying by r stretches it out to distance r.

e^(iθ) traces the unit circle as θ varies

The four green dots above are the special values you must commit to memory — they show up everywhere in HL exam questions. They’re just the four “compass points” of the unit circle:

θ = 0

e⁰ = 1

on the positive real axis

θ = π/2

e^iπ/2 = i

on the positive imaginary axis

θ = π

e^iπ = −1

on the negative real axis

Generalisation: e^2kπi = 1 for any integer k, because adding a full revolution (2π) brings you back to the starting point on the unit circle. Useful for simplifying powers like e^10πi = (e^2πi)⁵ = 1⁵ = 1.

Euler’s identity — the most beautiful equation

The special value e^iπ = −1 can be rearranged to give what many mathematicians call the most elegant equation ever written:

Euler’s Identity e^iπ + 1 = 0

Look at what’s in this single equation: e (the base of natural logarithms), i (the imaginary unit), π (the circle constant), 1 (the multiplicative identity), and 0 (the additive identity). Five fundamental constants from completely different parts of mathematics, linked by a single equation. That’s why people call it beautiful — and why it’s a regular feature of HL Paper 1 multiple-choice and short-answer questions.

Multiplication and division — index laws apply

Here’s the killer feature of exponential form. Because the i lives in the exponent, the rules of indices that you’ve used since GCSE work directly. That’s it — no special rules to memorise, just regular index manipulation.

Multiplication

r₁r₂e^{i(θ₁+θ₂)}

moduli multiply, exponents add (as always with indices)

Division

(r₁/r₂)e^{i(θ₁−θ₂)}

moduli divide, exponents subtract

Same rule as polar form, just expressed using e^iθ notation. If you’re confident with polar form, you already know how to multiply and divide in exponential form.

Powers in exponential form

This is where exponential form really earns its keep. To raise a complex number to a power, just use the index law (e^a)^b = e^ab:

Powers (re^iθ)ⁿ = rⁿe^inθ

So squaring multiplies the modulus by itself and doubles the argument. Cubing cubes the modulus and triples the argument. And so on. This pattern is essentially De Moivre’s theorem, which you’ll meet in detail in the next few notes.

Negative and fractional powers: the same rule works. (re^iθ)⁻¹ = (1/r)e^−iθ, and (re^iθ)^1/2 = √r · e^iθ/2. This one-line approach is dramatically faster than expanding (a+bi)ⁿ for large n.

Worked examples

WE 1

Convert Cartesian to exponential form

Express z = 2 + 2i in exponential (Euler’s) form, with the argument in the range −π < θ ≤ π.

Step 1: Sketch — z is in Q1 argument will be positive acute Step 2: Modulus by Pythagoras |z| = √(2² + 2²) = √8 = 2√2 Step 3: Argument arg(z) = arctan(2/2) = arctan(1) = π/4 Step 4: Write in exponential form z = 2√2 · e^(iπ/4) just bundle the modulus and argument into r·e^(iθ) — same numbers as polar form

WE 2

Multiplication in exponential form

Let z₁ = 4e^iπ/3 and z₂ = 5e^iπ/6. Find z₁z₂ in exponential form.

Step 1: Multiply the moduli 4 · 5 = 20 Step 2: Add the exponents (using index laws) e^(iπ/3) · e^(iπ/6) = e^(i(π/3 + π/6)) π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2 Step 3: Combine z₁z₂ = 20 · e^(iπ/2) notice this equals 20i, since e^(iπ/2) = i — the geometric meaning is “rotate to imaginary axis”

WE 3

Division in exponential form

Let z₁ = 18e^i(5π/6) and z₂ = 6e^i(π/4). Find z₁/z₂ in exponential form, with the argument in the range −π < θ ≤ π.

Step 1: Divide moduli 18 / 6 = 3 Step 2: Subtract exponents 5π/6 − π/4 = 10π/12 − 3π/12 = 7π/12 Step 3: Check range and write answer 7π/12 is in (−π, π] ✓ z₁/z₂ = 3 · e^(i·7π/12) cleaner than dividing in Cartesian form — no conjugate needed

WE 4

A power of a complex number

Given z = 3e^iπ/5, find z⁴ in exponential form, with the argument in the range −π < θ ≤ π.

Step 1: Apply the power rule (index laws) z⁴ = (3e^(iπ/5))⁴ = 3⁴ · e^(i·4π/5) Step 2: Compute the modulus 3⁴ = 81 Step 3: Check the argument is in range 4π/5 ≈ 2.513, π ≈ 3.142 → in range ✓ z⁴ = 81 · e^(i·4π/5) same calculation in Cartesian would mean expanding (a+bi)⁴ — this is dramatically faster

WE 5

Using the special values

Simplify each of the following:
(a) e^5πi (b) 4e^iπ + 4 (c) e^i·9π/2

Part (a): split off multiples of 2π e^(5πi) = e^(4πi) · e^(πi) = 1 · (−1) (a) e^(5πi) = −1 Part (b): use Euler’s identity e^(iπ) = −1 4 · e^(iπ) + 4 = 4(−1) + 4 = −4 + 4 = 0 (b) 4e^(iπ) + 4 = 0 Part (c): split off 4π 9π/2 = 4π + π/2 e^(i·9π/2) = e^(4πi) · e^(iπ/2) = 1 · i (c) e^(i·9π/2) = i always strip out multiples of 2π (or 2πi) — they vanish, leaving a small angle

WE 6

Combined operations

Let z₁ = 2e^iπ/4 and z₂ = 3e^iπ/12. Find (z₁)³ · z₂ in exponential form.

Step 1: Cube z₁ first using the power rule z₁³ = 2³ · e^(i·3π/4) = 8 · e^(i·3π/4) Step 2: Multiply by z₂ moduli: 8 · 3 = 24 exponents add: 3π/4 + π/12 common denominator 12: 9π/12 + π/12 = 10π/12 = 5π/6 Step 3: Check range and combine 5π/6 is in (−π, π] ✓ z₁³ · z₂ = 24 · e^(i·5π/6) complex number arithmetic at this level is just index laws + a check on the range — no expanding needed

💡 Top tips

Reach for exponential form whenever powers, roots, or repeated multiplication appear. It’s the fastest of the three forms for those operations.
Memorise the four compass-point values: e⁰ = 1, e^iπ/2 = i, e^iπ = −1, e^i3π/2 = −i. They show up everywhere.
Strip out full revolutions before simplifying. e^{i(θ + 2kπ)} = e^iθ for any integer k, so always reduce big arguments mod 2π first.
Standard index laws apply. e^a · e^b = e^a+b, e^a/e^b = e^a−b, (e^a)ⁿ = e^na. No special complex-number rules needed.
The position of i in the exponent doesn’t matter. e^iθ, e^θi, e^(θ)i — all the same number. Use whichever notation feels cleaner.
To convert exponential to Cartesian, go through polar first: re^iθ = r(cos θ + i sin θ), then expand. There’s no direct shortcut.
Many questions use the range 0 ≤ θ < 2π for exponential form (rather than −π < θ ≤ π). Read the question carefully.

⚠ Common mistakes

Trying to convert e^iθ to Cartesian form directly. You need to go through polar form first (use Euler’s formula to write it as cos θ + i sin θ).
Forgetting that the modulus must be positive. If you ever derive a “negative r“, absorb the minus into the angle by adding π — moduli are never negative.
Misapplying index laws. Don’t write e^iθ + e^iφ = e^i(θ+φ) — that’s only true for products, not sums.
Multiplying the exponents instead of adding. e^iθ₁ · e^iθ₂ = e^{i(θ₁+θ₂)}, not e^{i(θ₁θ₂)}.
Forgetting i² = −1 doesn’t apply directly here. The whole point of exponential form is that you don’t need to track i² separately — index laws on the exponent take care of it.
Using degrees instead of radians. All complex-number arguments in HL are in radians. Always.
Forgetting to strip 2π before simplifying. e^i·17π looks complicated until you notice 17π = 8·(2π) + π, so it equals e^iπ = −1.

Exponential form is genuinely the most elegant of the three notations — and once you’ve got it under your fingers, you’ll use it for nearly every advanced complex-number problem. The next note ties all three forms together: Cartesian, polar, and exponential, and shows you how to flip between them as needed. After that comes the truly powerful stuff: De Moivre’s theorem, complex roots, and roots of unity. Exponential form is the engine that makes those topics tractable.

Need help with Exponential Form?

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

Book Free Session →