Core Concept
A Complex Number is an extension of real numbers, taking the form z = a + b i z = a + bi z = a + bi where a , b a, b a , b are real numbers and i i i is the imaginary unit.
Imaginary Unit
The imaginary unit i i i satisfies i 2 = − 1 i^2 = -1 i 2 = − 1 .
i 2 = − 1 , i = − 1 i^2 = -1, \quad i = \sqrt{-1} i 2 = − 1 , i = − 1
Powers of i i i :
i 1 = i i^1 = i i 1 = i
i 2 = − 1 i^2 = -1 i 2 = − 1
i 3 = − i i^3 = -i i 3 = − i
i 4 = 1 i^4 = 1 i 4 = 1
i 4 k + r = i r i^{4k+r} = i^r i 4 k + r = i r (k ∈ Z , r ∈ { 0 , 1 , 2 , 3 } k \in \mathbb{Z}, r \in \{0,1,2,3\} k ∈ Z , r ∈ { 0 , 1 , 2 , 3 } )
Form of Complex Numbers
z = a + b i z = a + bi z = a + bi
Where:
a a a is the real part , denoted Re ( z ) \text{Re}(z) Re ( z )
b b b is the imaginary part , denoted Im ( z ) \text{Im}(z) Im ( z )
When b = 0 b = 0 b = 0 , z z z is a real number
When a = 0 , b ≠ 0 a = 0, b \neq 0 a = 0 , b = 0 , z z z is a pure imaginary number
When b ≠ 0 b \neq 0 b = 0 , z z z is an imaginary number
Complex Equality
a + b i = c + d i ⇔ a = c and b = d a + bi = c + di \Leftrightarrow a = c \text{ and } b = d a + bi = c + d i ⇔ a = c and b = d
Complex Plane
Geometric Representation
Complex number z = a + b i z = a + bi z = a + bi can be represented as point ( a , b ) (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 + b i z = a + bi z = a + bi can also be viewed as vector O Z → \overrightarrow{OZ} OZ from origin O O O to point ( a , b ) (a, b) ( a , b ) .
Modulus of Complex Numbers
Definition
The modulus of complex number z = a + b i z = a + bi z = a + bi , denoted ∣ z ∣ |z| ∣ z ∣ :
∣ z ∣ = ∣ a + b i ∣ = a 2 + b 2 |z| = |a + bi| = \sqrt{a^2 + b^2} ∣ z ∣ = ∣ a + bi ∣ = a 2 + b 2
Geometric Meaning
∣ z ∣ |z| ∣ z ∣ represents the distance from point z z z to the origin in the complex plane.
Properties
∣ z ∣ ≥ 0 |z| \geq 0 ∣ z ∣ ≥ 0 , with equality iff z = 0 z = 0 z = 0
∣ z 1 ⋅ z 2 ∣ = ∣ z 1 ∣ ⋅ ∣ z 2 ∣ |z_1 \cdot z_2| = |z_1| \cdot |z_2| ∣ z 1 ⋅ z 2 ∣ = ∣ z 1 ∣ ⋅ ∣ z 2 ∣
∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} z 2 z 1 = ∣ z 2 ∣ ∣ z 1 ∣ (z 2 ≠ 0 z_2 \neq 0 z 2 = 0 )
∣ z 1 + z 2 ∣ ≤ ∣ z 1 ∣ + ∣ z 2 ∣ |z_1 + z_2| \leq |z_1| + |z_2| ∣ z 1 + z 2 ∣ ≤ ∣ z 1 ∣ + ∣ z 2 ∣ (triangle inequality)
Conjugate
Definition
The conjugate of complex number z = a + b i z = a + bi z = a + bi , denoted z ˉ \bar{z} z ˉ :
z ˉ = a − b i \bar{z} = a - bi z ˉ = a − bi
Geometric Meaning
z ˉ \bar{z} z ˉ is the reflection of z z z across the real axis.
Properties
z 1 ± z 2 ‾ = z 1 ˉ ± z 2 ˉ \overline{z_1 \pm z_2} = \bar{z_1} \pm \bar{z_2} z 1 ± z 2 = z 1 ˉ ± z 2 ˉ
z 1 ⋅ z 2 ‾ = z 1 ˉ ⋅ z 2 ˉ \overline{z_1 \cdot z_2} = \bar{z_1} \cdot \bar{z_2} z 1 ⋅ z 2 = z 1 ˉ ⋅ z 2 ˉ
z ⋅ z ˉ = ∣ z ∣ 2 = a 2 + b 2 z \cdot \bar{z} = |z|^2 = a^2 + b^2 z ⋅ z ˉ = ∣ z ∣ 2 = a 2 + b 2
z + z ˉ = 2 a = 2 Re ( z ) z + \bar{z} = 2a = 2\text{Re}(z) z + z ˉ = 2 a = 2 Re ( z )
z − z ˉ = 2 b i = 2 i Im ( z ) z - \bar{z} = 2bi = 2i\text{Im}(z) z − z ˉ = 2 bi = 2 i Im ( z )
Operations
Addition and Subtraction
( a + b i ) ± ( c + d i ) = ( a ± c ) + ( b ± d ) i (a + bi) \pm (c + di) = (a \pm c) + (b \pm d)i ( a + bi ) ± ( c + d i ) = ( a ± c ) + ( b ± d ) i
Multiplication
( a + b i ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i (a + bi)(c + di) = (ac - bd) + (ad + bc)i ( a + bi ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i
Division
a + b i c + d i = ( a + b i ) ( c − d i ) ( c + d i ) ( c − d i ) = ( a c + b d ) + ( b c − a d ) i c 2 + d 2 \frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} c + d i a + bi = ( c + d i ) ( c − d i ) ( a + bi ) ( c − d i ) = c 2 + d 2 ( a c + b d ) + ( b c − a d ) i
Method : Multiply numerator and denominator by conjugate of denominator
CSCA Practice Problems
[Example 1] Basic (Difficulty ★★☆☆☆)
Given complex number z = 3 + 4 i z = 3 + 4i z = 3 + 4 i , find ∣ z ∣ |z| ∣ z ∣ and z ˉ \bar{z} z ˉ .
Solution :
Modulus :
∣ z ∣ = 3 2 + 4 2 = 25 = 5 |z| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 ∣ z ∣ = 3 2 + 4 2 = 25 = 5
Conjugate :
z ˉ = 3 − 4 i \bar{z} = 3 - 4i z ˉ = 3 − 4 i
Answer : ∣ z ∣ = 5 |z| = 5 ∣ z ∣ = 5 , z ˉ = 3 − 4 i \bar{z} = 3 - 4i z ˉ = 3 − 4 i
[Example 2] Intermediate (Difficulty ★★★☆☆)
Calculate ( 2 + 3 i ) ( 1 − 2 i ) (2 + 3i)(1 - 2i) ( 2 + 3 i ) ( 1 − 2 i ) .
Solution :
( 2 + 3 i ) ( 1 − 2 i ) = 2 − 4 i + 3 i − 6 i 2 (2 + 3i)(1 - 2i) = 2 - 4i + 3i - 6i^2 ( 2 + 3 i ) ( 1 − 2 i ) = 2 − 4 i + 3 i − 6 i 2
= 2 − i + 6 = 8 − i = 2 - i + 6 = 8 - i = 2 − i + 6 = 8 − i
Answer : 8 − i 8 - i 8 − i
Common Misconceptions
❌ Misconception 1: Treating i i i as a variable
Wrong : Thinking i i i can be further simplified like algebraic variables
Correct : i i i is the imaginary unit with i 2 = − 1 i^2 = -1 i 2 = − 1 , not a variable
❌ Misconception 2: Wrong modulus calculation
Wrong : ∣ 3 + 4 i ∣ = 3 + 4 = 7 |3 + 4i| = 3 + 4 = 7 ∣3 + 4 i ∣ = 3 + 4 = 7
Correct : ∣ 3 + 4 i ∣ = 3 2 + 4 2 = 5 |3 + 4i| = \sqrt{3^2 + 4^2} = 5 ∣3 + 4 i ∣ = 3 2 + 4 2 = 5
❌ Misconception 3: Wrong conjugate sign
Wrong : 3 + 4 i ‾ = − 3 − 4 i \overline{3 + 4i} = -3 - 4i 3 + 4 i = − 3 − 4 i
Correct : 3 + 4 i ‾ = 3 − 4 i \overline{3 + 4i} = 3 - 4i 3 + 4 i = 3 − 4 i (only imaginary part changes sign)
Study Tips
✅ Understand imaginary unit : i 2 = − 1 i^2 = -1 i 2 = − 1 is fundamental
✅ Master operations : Addition, subtraction, multiplication, division
✅ Remember modulus and conjugate : Their geometric meanings and properties
✅ Practice division : Rationalizing denominator is key
✅ 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.