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complex-numbersDifficulty: Basiccomplex-numbersalgebra
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复数fùshù

complex number
5 min read
Updated November 2, 2025
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Core Concept

A Complex Number is an extension of real numbers, taking the form z=a+biz = a + bi where a,ba, b are real numbers and ii is the imaginary unit.

Imaginary Unit

The imaginary unit ii satisfies i2=−1i^2 = -1.

i2=−1,i=−1i^2 = -1, \quad i = \sqrt{-1}

Powers of ii:

  • i1=ii^1 = i
  • i2=−1i^2 = -1
  • i3=−ii^3 = -i
  • i4=1i^4 = 1
  • i4k+r=iri^{4k+r} = i^r (k∈Z,r∈{0,1,2,3}k \in \mathbb{Z}, r \in \{0,1,2,3\})

Form of Complex Numbers

z=a+biz = a + bi

Where:

  • aa is the real part, denoted Re(z)\text{Re}(z)
  • bb is the imaginary part, denoted Im(z)\text{Im}(z)
  • When b=0b = 0, zz is a real number
  • When a=0,b≠0a = 0, b \neq 0, zz is a pure imaginary number
  • When b≠0b \neq 0, zz is an imaginary number

Complex Equality

a+bi=c+di⇔a=c and b=da + bi = c + di \Leftrightarrow a = c \text{ and } b = d

Complex Plane

Geometric Representation

Complex number z=a+biz = a + bi can be represented as point (a,b)(a, b) in the complex plane:

  • Horizontal axis (real axis): Represents real part
  • Vertical axis (imaginary axis): Represents imaginary part

Vector Representation

Complex number z=a+biz = a + bi can also be viewed as vector OZ→\overrightarrow{OZ} from origin OO to point (a,b)(a, b).

Modulus of Complex Numbers

Definition

The modulus of complex number z=a+biz = a + bi, denoted ∣z∣|z|:

∣z∣=∣a+bi∣=a2+b2|z| = |a + bi| = \sqrt{a^2 + b^2}

Geometric Meaning

∣z∣|z| represents the distance from point zz to the origin in the complex plane.

Properties

  1. ∣z∣≥0|z| \geq 0, with equality iff z=0z = 0
  2. ∣z1⋅z2∣=∣z1∣⋅∣z2∣|z_1 \cdot z_2| = |z_1| \cdot |z_2|
  3. ∣z1z2∣=∣z1∣∣z2∣\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} (z2≠0z_2 \neq 0)
  4. ∣z1+z2∣≤∣z1∣+∣z2∣|z_1 + z_2| \leq |z_1| + |z_2| (triangle inequality)

Conjugate

Definition

The conjugate of complex number z=a+biz = a + bi, denoted zˉ\bar{z}:

zˉ=a−bi\bar{z} = a - bi

Geometric Meaning

zˉ\bar{z} is the reflection of zz across the real axis.

Properties

  1. z1±z2‾=z1ˉ±z2ˉ\overline{z_1 \pm z_2} = \bar{z_1} \pm \bar{z_2}
  2. z1⋅z2‾=z1ˉ⋅z2ˉ\overline{z_1 \cdot z_2} = \bar{z_1} \cdot \bar{z_2}
  3. z⋅zˉ=∣z∣2=a2+b2z \cdot \bar{z} = |z|^2 = a^2 + b^2
  4. z+zˉ=2a=2Re(z)z + \bar{z} = 2a = 2\text{Re}(z)
  5. z−zˉ=2bi=2iIm(z)z - \bar{z} = 2bi = 2i\text{Im}(z)

Operations

Addition and Subtraction

(a+bi)±(c+di)=(a±c)+(b±d)i(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)i

Multiplication

(a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

Division

a+bic+di=(a+bi)(c−di)(c+di)(c−di)=(ac+bd)+(bc−ad)ic2+d2\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}

Method: Multiply numerator and denominator by conjugate of denominator

CSCA Practice Problems

[Example 1] Basic (Difficulty ★★☆☆☆)

Given complex number z=3+4iz = 3 + 4i, find ∣z∣|z| and zˉ\bar{z}.

Solution:

Modulus: ∣z∣=32+42=25=5|z| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

Conjugate: zˉ=3−4i\bar{z} = 3 - 4i

Answer: ∣z∣=5|z| = 5, zˉ=3−4i\bar{z} = 3 - 4i


[Example 2] Intermediate (Difficulty ★★★☆☆)

Calculate (2+3i)(1−2i)(2 + 3i)(1 - 2i).

Solution:

(2+3i)(1−2i)=2−4i+3i−6i2(2 + 3i)(1 - 2i) = 2 - 4i + 3i - 6i^2 =2−i+6=8−i= 2 - i + 6 = 8 - i

Answer: 8−i8 - i

Common Misconceptions

❌ Misconception 1: Treating ii as a variable

Wrong: Thinking ii can be further simplified like algebraic variables

Correct: ii is the imaginary unit with i2=−1i^2 = -1, not a variable

❌ Misconception 2: Wrong modulus calculation

Wrong: ∣3+4i∣=3+4=7|3 + 4i| = 3 + 4 = 7

Correct: ∣3+4i∣=32+42=5|3 + 4i| = \sqrt{3^2 + 4^2} = 5

❌ Misconception 3: Wrong conjugate sign

Wrong: 3+4i‾=−3−4i\overline{3 + 4i} = -3 - 4i

Correct: 3+4i‾=3−4i\overline{3 + 4i} = 3 - 4i (only imaginary part changes sign)

Study Tips

  1. ✅ Understand imaginary unit: i2=−1i^2 = -1 is fundamental
  2. ✅ Master operations: Addition, subtraction, multiplication, division
  3. ✅ Remember modulus and conjugate: Their geometric meanings and properties
  4. ✅ Practice division: Rationalizing denominator is key
  5. ✅ Understand geometry: Points and vectors in complex plane

💡 Exam Tip: Complex numbers are important in high school math. Relatively straightforward in CSCA exams, but basic operations and concepts must be mastered! Accounts for about 10-15% of algebra problems.

How the real exam tests it

Questions from real CSCA papers that test this term. Work each one out before you open the answer.

  1. CSCA Math April 2026 Past Paper
    (z−2i)(2−i)=5( z - 2 i ) ( 2 - i ) = 5, then z=z =
    • A.2+3i2 + 3 i
    • B.2−3i2 - 3 i
    • C.3−2i3 - 2 i
    • D.3+2i3 + 2 i
    Show answer & solution

    Answer: A

    z−2i=52−i=5(2+i)5=2+iz - 2 i = \frac { 5 } { 2 - i } = \frac { 5 ( 2 + i ) } { 5 } = 2 + i. So z=2+i+2i=2+3iz = 2 + i + 2 i = 2 + 3 i.
    Take the full CSCA Math April 2026 Past Paper — timed & graded
  2. CSCA Math June 2026 Past Paper
    Given a complex number z=−1+iz = -1 + i, where ii is the imaginary unit, then z+1z+2+z+2z+1=\frac{z+1}{z+2} + \frac{z+2}{z+1} =
    • A.32−12i\frac{3}{2} - \frac{1}{2}i
    • B.32+12i\frac{3}{2} + \frac{1}{2}i
    • C.12−32i\frac{1}{2} - \frac{3}{2}i
    • D.1−i1 - i
    Show answer & solution

    Answer: A

    z+1=iz + 1 = i and z+2=1+iz + 2 = 1 + i. Then z+1z+2=i1+i=i(1−i)(1+i)(1−i)=1+i2=12+12i\frac{z+1}{z+2} = \frac{i}{1+i} = \frac{i(1-i)}{(1+i)(1-i)} = \frac{1+i}{2} = \frac{1}{2} + \frac{1}{2}i, and z+2z+1=1+ii=(1+i)(−i)i⋅(−i)=1−i\frac{z+2}{z+1} = \frac{1+i}{i} = \frac{(1+i)(-i)}{i \cdot (-i)} = 1 - i. Adding the two gives 32−12i\frac{3}{2} - \frac{1}{2}i. Choose (A).
    Take the full CSCA Math June 2026 Past Paper — timed & graded

Practice Exercises

Practice to reinforce your understanding