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NotesMath AI HLTopic 1.12Complex numbers in Cartesian form
Back to Math AI HL Topics
1.12.12 min read

Complex numbers in Cartesian form

IB Mathematics: Applications and Interpretation • Unit 1

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Contents

  • What is a complex number? Add, subtract, multiply
  • Conjugate, division, the Argand diagram & modulus
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.
The defining rule and the Cartesian (rectangular) form.
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

  1. Add real parts (3 + 1) and imaginary parts (2 + (−5)).
  2. 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

  1. Expand all four products.
  2. Simplify, keeping the i² term.
  3. 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.
Conjugate, and why it produces a real number.
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

  1. Multiply top and bottom by the conjugate of the bottom, 2 + i.
  2. Top: (5 + i)(2 + i) = 10 + 5i + 2i + i² = 10 + 7i − 1 = 9 + 7i.
  3. Bottom: (2 − i)(2 + i) = 2² + 1² = 5 (real, as promised).
  4. 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.
Modulus = distance from the origin on the Argand diagram.

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

  1. Real part across, imaginary part up.
  2. Modulus by Pythagoras.
  3. 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

  1. Use the quadratic formula with a = 1, b = −4, c = 13.
  2. The discriminant is negative, so the roots are complex.
  3. 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).

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Given z₁ = 4 − 3i and z₂ = −1 + 2i, find z₁ + z₂ and z₁ − z₂. [2 marks]

Related Math AI HL Topics

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1.1.1Converting to standard form
1.1.2Back to ordinary form
1.1.3Calculations with standard form
1.1.4Validity checks and GDC output
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