i is the number whose square is −1: Some equations, like x² = −1, have no real solution — no real number squares to a negative.
So mathematicians invented a new number i with the single rule i² = −1.
A complex number is then written z = a + bi, where a is the real part and b is the imaginary part (both are ordinary real numbers).
This is exactly how engineers describe alternating currents and how physicists handle waves — i is a genuine tool, not a trick.
Add and subtract: just collect like terms: Treat i like a letter (an x): add the real parts together and the imaginary parts together. Nothing new to memorise.
IB-style question — add and subtract
Let z₁ = 3 + 2i and z₂ = 1 − 5i.
Find z₁ + z₂ and z₁ − z₂.
Step by step
- Add real parts (3 + 1) and imaginary parts (2 + (−5)).
- Subtract: real (3 − 1), imaginary (2 − (−5)).
Final answer
z₁ + z₂ = 4 − 3i and z₁ − z₂ = 2 + 7i.
Multiply: expand, then replace i² with −1: Multiply out the brackets exactly like (a + b)(c + d). The only new step is that whenever you get i², swap it for −1.
IB-style question — multiply
Find (2 + 3i)(1 − 4i).
Step by step
- Expand all four products.
- Simplify, keeping the i² term.
- Replace i² with −1, so −12i² = +12.
Final answer
(2 + 3i)(1 − 4i) = 14 − 5i.
The conjugate flips the sign of the imaginary part: The conjugate of z = a + bi is z* = a − bi (some books write z̄).
Its magic property: z × z\* is always a real number, because
(a + bi)(a − bi) = a² − (bi)² = a² + b² — the i's cancel.
That is exactly what lets us divide complex numbers.
To divide: multiply top and bottom by the conjugate of the bottom: We can't divide by 'i' directly, so we clear the i from the denominator — just like rationalising a surd. Multiply the fraction by (conjugate of the bottom)/(conjugate of the bottom). The bottom becomes a real number and you split into real + imaginary parts.
IB-style question — divide
Express (5 + i)/(2 − i) in the form a + bi.
Step by step
- Multiply top and bottom by the conjugate of the bottom, 2 + i.
- Top: (5 + i)(2 + i) = 10 + 5i + 2i + i² = 10 + 7i − 1 = 9 + 7i.
- Bottom: (2 − i)(2 + i) = 2² + 1² = 5 (real, as promised).
- Split into real + imaginary parts.
Final answer
(5 + i)/(2 − i) = 9/5 + 7/5 i.
Picture a complex number on the Argand diagram: An Argand diagram plots z = a + bi as the point (a, b) — real part across, imaginary part up.
The modulus |z| is then simply the distance from the origin to that point, found by Pythagoras:
|z| = √(a² + b²). It measures the 'size' of the complex number.
IB-style question — plot and find the modulus
z = 3 + 4i is plotted on an Argand diagram.
State its coordinates and find |z|.
Step by step
- Real part across, imaginary part up.
- Modulus by Pythagoras.
- Simplify.
Final answer
z is the point (3, 4); |z| = 5 (a 3–4–5 right triangle).
IB-style question — a quadratic with complex roots
Solve the equation z² − 4z + 13 = 0, giving your answers in the form a + bi.
Step by step
- Use the quadratic formula with a = 1, b = −4, c = 13.
- The discriminant is negative, so the roots are complex.
- Write √(−36) = 6i and simplify.
Final answer
z = 2 + 3i or z = 2 − 3i — a complex-conjugate pair (real-coefficient quadratics always give conjugate roots).