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Complex Numbers

The complex plane, modulus and argument, the four operations, conjugates, and quadratic roots.

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lessons in this chapter

previous exam questions

3 hr

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Hard

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Last reviewed Sep 2026 · GetCSCA editors

1. Complex Numbers

Study order

Work the lessons top to bottom — each ends with practice, a timed set and a 5-stage drill. Then prove it in a full 48Q mock.

Complex Numbers

Lesson 1 · target ~70s per question

a + bi arithmetic, conjugates, modulus, division by rationalizing, and complex roots of quadratics.

Why CSCA tests this

Complex numbers test conjugate discipline: division, modulus, and quadratic roots with Δ < 0 all run through multiply-by-conjugate or the conjugate-pair theorem.

Key points — memorize

Write every number as a + bi and operate component-wise for addition/subtraction; multiply with FOIL then replace i² by −1.
Divide by multiplying top and bottom by the conjugate of the denominator, using (a+bi)(a−bi) = a² + b² to real-ize it.
Modulus |a + bi| = √(a² + b²) is the origin distance, and |z|² = z·z̄ converts modulus conditions into real equations.
Reduce powers of i modulo 4 (i, −1, −i, 1 cycle) before any other algebra.
Real-coefficient quadratics with Δ < 0 have conjugate roots (−b ± i√|Δ|)/2a — if one root is given, the other is its conjugate.

Formula sheet

i² = −1
(a + bi)(a − bi) = a² + b²
|a + bi| = √(a² + b²)
1/i = −i
Δ < 0 ⇒ x = (−b ± i√|Δ|)/2a
|z|² = z·z̄; z̄ = a − bi for z = a + bi
i-cycle: i¹ = i, i² = −1, i³ = −i, i⁴ = 1 (period 4)
z₁ + z₂ = (a+c) + (b+d)i; z₁z₂ via FOIL with i² = −1

Classic traps

Leaving i² un-replaced after FOIL, or replacing it with 1 instead of −1.
Rationalizing with the wrong conjugate (same sign) so the denominator stays complex.
Forgetting to divide both real and imaginary parts after rationalizing.
Reporting only one complex root of a real quadratic instead of the conjugate pair.

Exam tactic

On any complex division, write the conjugate multiplier explicitly on the first line — the denominator becomes a² + b² mechanically.

Worked examples

Example 1: Compute (2 + 3i)(1 − i).
1. FOIL: 2 − 2i + 3i − 3i² = 2 + i − 3(−1).
2. Simplify: 5 + i.
Answer: 5 + i
Example 2: Compute (1 + i)/(1 − i).
1. Multiply by conjugate (1 + i): (1 + i)²/(1 + 1) = 2i/2.
Answer: i

FoundationPractice →BasicPractice →CSCA StandardPractice →Timed set · ~70s/Q →

Previous exam questions

Real CSCA-style questions tagged to Complex Numbers — answer right here.

3 previous exam questions · 0/0 correct

Dec 2025 · Q46complex-numbers
$z^3 = 1$ , $z \ne 1$ : $1+z+\cdots+z^{999} =$
Jan 2026 · Q45complex-numbers
$z$ on line $x-y=0$ , root of $x^2+mx+4=0$ : $m =$
Apr 2026 · Q46complex-numbers
$(z-2i)(2-i) = 5$ : $z =$

Related guides

30-day study plan → fit this chapter into a week-by-week system
75-second pacing rule → hold 75s/question once the content clicks
Top 10 mistakes → the error patterns that cost the most points

1 lessons · one chapter

Finished Complex Numbers? Prove it.

Run a timed set, clear every mistake, then take a full 48Q mock.

Practice questions →Full mock →