IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 ~10 min read

Operations with Complex Numbers

Now that you know what a complex number is, the natural next question is: how do you do arithmetic with them? Adding and subtracting are easy — just keep real with real and imaginary with imaginary. Multiplying needs you to expand brackets and remember that i² = −1. Dividing is the trickiest, but a clever trick involving the complex conjugate turns it into something you can do every time. Get these four operations under your fingers and you’ll be ready for everything else in the topic — Argand diagrams, polar form, De Moivre’s theorem — they all rest on the basic arithmetic you’ll learn here.

📘 What you need to know

To add or subtract: combine the real parts, then combine the imaginary parts. Treat them like separate variables.
To multiply: expand the brackets like normal binomials, then use i² = −1 to simplify.
The complex conjugate of z = a + bi is z* = a − bi (just flip the sign of the imaginary part).
z · z* is always a real number — equal to a² + b². This is the key to division.
To divide: multiply both top and bottom by the conjugate of the denominator. The denominator becomes real, and you can split into real and imaginary parts.
Powers of i cycle every four: i, −1, −i, 1, i, −1, −i, 1, … Use index laws and the cycle for higher powers.
Your GDC has a complex mode (sometimes called “rectangular” or “a + bi”). Use it for fast arithmetic checks, especially on Paper 2.

Adding and subtracting complex numbers

This part is the easiest. To add two complex numbers, just add the real parts and add the imaginary parts — separately. Same for subtraction.

Addition and subtraction (a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) − (c + di) = (a − c) + (b − d)i

For example: (5 + 3i) + (2 + 7i) = (5 + 2) + (3 + 7)i = 7 + 10i. And (5 + 3i) − (2 + 7i) = (5 − 2) + (3 − 7)i = 3 − 4i. Real with real, imaginary with imaginary. Don’t mix.

Treat i like an algebraic letter for now (even though it’s a number). You’d never combine 5x with 7 — they’re different types of thing. Same with the real and imaginary parts: keep them in their own lanes.

Multiplying complex numbers

To multiply two complex numbers, expand the brackets exactly like you would for any pair of binomials. The only twist comes at the end: whenever you see i², replace it with −1.

Multiplication (a + bi)(c + di) = ac + adi + bci + bdi²
= (ac − bd) + (ad + bc)i

You don’t have to memorise this expanded form — just remember the steps: expand, replace i² with −1, then collect the real and imaginary parts.

Worked through: (3 + 2i)(4 + 5i) = 12 + 15i + 8i + 10i². Now replace i² with −1: 12 + 15i + 8i + 10(−1) = 12 + 23i − 10 = 2 + 23i.

Powers of i — the cycle of four

Powers of i look intimidating until you spot the pattern. Just compute the first few and the rest become obvious.

The four-step cycle of powers of i

So to compute any power of i, just find the remainder when the power is divided by 4 — that tells you which point on the cycle you’ve landed on:

Remainder when dividing by 4: remainder 0 → 1, remainder 1 → i, remainder 2 → −1, remainder 3 → −i.

🤔 Why does the cycle work?

Every time you multiply by i, you’re “moving forward” by one step around the cycle. Since i⁴ = (i²)² = (−1)² = 1, multiplying by i four times brings you back to where you started. So the powers can never escape this four-step loop.

The complex conjugate

For every complex number, there’s a partner called its conjugate. You get the conjugate by flipping the sign of the imaginary part.

Complex conjugate if z = a + bi, then z* = a − bi

So the conjugate of 3 + 5i is 3 − 5i. The conjugate of −2 − 7i is −2 + 7i. The real part stays put; only the sign of the imaginary part changes. Three useful properties to remember:

z + z*

always real

imaginary parts cancel

z − z*

always imaginary

real parts cancel

z · z*

a² + b²

always real and positive

The third one — that z · z* is always real — is the magic trick that makes division work. When we multiply (a + bi)(a − bi), the difference-of-two-squares pattern gives a² − (bi)² = a² − b²·(−1) = a² + b². The i has vanished.

Dividing complex numbers

Dividing one complex number by another doesn’t have a clean rule the way addition or multiplication do. Instead, there’s a neat trick that turns division into multiplication. The idea is to “rationalise” the denominator — much like rationalising surds in earlier years.

🧭 Recipe — dividing complex numbers

Write the division as a fraction. So (a + bi) ÷ (c + di) becomes (a + bi) / (c + di).
Multiply top and bottom by the conjugate of the denominator — that’s c − di.
Expand both numerator and denominator. Use i² = −1 to simplify.
The denominator should now be a real number (specifically, c² + d²).
Split the fraction into real and imaginary parts to give the final answer in Cartesian form.

Let me show you the move with a quick example. Compute (4 + 2i) / (1 − i):

Worked through 4 + 2i1 − i · 1 + i1 + i = (4 + 2i)(1 + i)(1 − i)(1 + i) = 4 + 4i + 2i + 2i²1 − i² = 2 + 6i2 = 1 + 3i

Why it works: multiplying the denominator by its conjugate turns it into a real number, and dividing by a real number is just splitting the fraction into real and imaginary halves. The conjugate trick is to division what “rationalising the denominator” was for surds.

Worked examples

WE 1

Add and subtract complex numbers

Let z₁ = 7 − 2i and z₂ = −3 + 5i. Find:
(a) z₁ + z₂ (b) z₁ − z₂

Part (a): add real parts, add imaginary parts z₁ + z₂ = (7 + (−3)) + (−2 + 5)i = 4 + 3i (a) z₁ + z₂ = 4 + 3i Part (b): subtract — keep brackets careful with signs z₁ − z₂ = (7 − (−3)) + (−2 − 5)i = 10 − 7i (b) z₁ − z₂ = 10 − 7i subtraction is where sign mistakes love to happen — a minus times a minus is a plus

WE 2

Simplify with multipliers

Simplify the expression 3(4 + 2i) − 2(1 − 5i).

Step 1: Distribute the multipliers 3(4 + 2i) = 12 + 6i 2(1 − 5i) = 2 − 10i Step 2: Subtract — careful with the signs (12 + 6i) − (2 − 10i) = 12 − 2 + 6i − (−10i) = 10 + 16i 3(4 + 2i) − 2(1 − 5i) = 10 + 16i distribute first, THEN combine real with real and imaginary with imaginary

WE 3

Multiply two complex numbers

Find the value of (5 + 2i)(3 − 4i).

Step 1: Expand the brackets like a normal binomial product (5 + 2i)(3 − 4i) = 15 − 20i + 6i − 8i² Step 2: Replace i² with −1 = 15 − 20i + 6i − 8(−1) = 15 − 20i + 6i + 8 Step 3: Collect real and imaginary parts real: 15 + 8 = 23 imaginary: −20 + 6 = −14 (5 + 2i)(3 − 4i) = 23 − 14i the i² term flipped sign and joined the real part — the most important step

WE 4

Conjugate properties

Let z = 4 + 3i. Find z*, then verify that z · z* is real.

Step 1: Write down the conjugate by flipping the imaginary sign z = 4 + 3i → z* = 4 − 3i Step 2: Compute z · z* using difference-of-squares pattern z · z* = (4 + 3i)(4 − 3i) = 4² − (3i)² (difference of squares) = 16 − 9i² = 16 − 9(−1) = 16 + 9 = 25 z · z* = 25 — a real number ✓ notice z · z* = 4² + 3² = 25 — exactly a² + b²

WE 5

Divide two complex numbers

Find (5 + 4i) ÷ (2 + 3i), giving your answer in Cartesian form.

Step 1: Write as a fraction (5 + 4i) / (2 + 3i) Step 2: Multiply top and bottom by conjugate of denominator (2 − 3i) = [(5 + 4i)(2 − 3i)] / [(2 + 3i)(2 − 3i)] Step 3: Expand the numerator (5 + 4i)(2 − 3i) = 10 − 15i + 8i − 12i² = 10 − 15i + 8i + 12 (i² = −1) = 22 − 7i Step 4: Simplify the denominator (2 + 3i)(2 − 3i) = 4 − 9i² = 4 + 9 = 13 Step 5: Split into Cartesian form = (22 − 7i) / 13 = 22/13 − (7/13)i (5 + 4i) ÷ (2 + 3i) = 22/13 − 7i/13 the conjugate trick makes the bottom real every single time — that’s why it always works

WE 6

Evaluate a high power of i

Find the value of i⁵⁰.

Step 1: Divide the power by 4 to find the remainder 50 ÷ 4 = 12 remainder 2 so 50 = 4·12 + 2 Step 2: Use index laws to split off the multiple of 4 i⁵⁰ = i^{4·12 + 2} = (i⁴)¹² · i² = 1¹² · i² (since i⁴ = 1) = 1 · (−1) = −1 i⁵⁰ = −1 the answer is whichever of {1, i, −1, −i} matches the remainder mod 4 — here remainder 2 → i² = −1

💡 Top tips

Treat i like a letter while you’re expanding, then replace i² with −1 at the end. This is the cleanest workflow.
The conjugate trick for division is non-negotiable. Multiplying both top and bottom by the conjugate of the denominator always gives a real denominator. Always.
For powers of i, find the remainder mod 4. Remainder 0 → 1, remainder 1 → i, remainder 2 → −1, remainder 3 → −i. That’s the whole rule.
Use the difference-of-squares pattern for z · z*. (a + bi)(a − bi) = a² + b². Faster than expanding the long way.
Watch the sign in the imaginary part. When subtracting, “minus a minus” gives a plus. Bracket your work to avoid slips.
Use your GDC’s complex mode for sanity checks on Paper 2. It will compute (5+2i)(3−4i) directly — but you still need to show working on Paper 1.
Final answer in Cartesian form. Even after a divide, write your answer as a + bi, not as a single fraction. Examiners want the real and imaginary parts separated.

⚠ Common mistakes

Forgetting i² = −1. Leaving an i² term in your final answer is the most common error in this topic. Replace it as soon as it appears.
Sign errors when subtracting. (3 + 2i) − (5 − 7i) = (3 − 5) + (2 − (−7))i = −2 + 9i. The negatives compound — go slow.
Multiplying by the wrong conjugate. When dividing, you multiply top and bottom by the conjugate of the denominator, not the numerator.
Stopping at i² without simplifying. i² isn’t a number — you need to write it as −1 before the final answer.
Computing iⁿ for big n the long way. Always reduce mod 4 first. i⁹⁷ isn’t worth multiplying out by hand.
Forgetting to multiply BOTH top and bottom by the conjugate. Multiplying just the bottom changes the value of the fraction. Both, every time.
Mis-identifying the conjugate. Conjugate of a − bi is a + bi (not a − (−b)i = a + bi — wait, that’s the same. Just flip the sign of the imaginary part, whatever it is).

Once you can confidently do all four operations on complex numbers, you’ve cleared the biggest hurdle in this topic. Everything that comes next — Argand diagrams, modulus and argument, polar form — builds on the same arithmetic, just dressed up in different notation. Spend an evening practising 10–15 mixed problems and you’ll be set. Up next: a geometric way to see complex numbers, called the Argand diagram.

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