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complex-numbers难度:基础complex-numbersalgebra
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complex number
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更新于 2025-11-02
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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 (kZ,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,b0a = 0, b \neq 0, zz is a pure imaginary number
  • When b0b \neq 0, zz is an imaginary number

Complex Equality

a+bi=c+dia=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. z0|z| \geq 0, with equality iff z=0z = 0
  2. z1z2=z1z2|z_1 \cdot z_2| = |z_1| \cdot |z_2|
  3. z1z2=z1z2\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} (z20z_2 \neq 0)
  4. z1+z2z1+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=abi\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. z1z2=z1ˉz2ˉ\overline{z_1 \cdot z_2} = \bar{z_1} \cdot \bar{z_2}
  3. zzˉ=z2=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. zzˉ=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)=(acbd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

Division

a+bic+di=(a+bi)(cdi)(c+di)(cdi)=(ac+bd)+(bcad)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=34i\bar{z} = 3 - 4i

Answer: z=5|z| = 5, zˉ=34i\bar{z} = 3 - 4i


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

Calculate (2+3i)(12i)(2 + 3i)(1 - 2i).

Solution:

(2+3i)(12i)=24i+3i6i2(2 + 3i)(1 - 2i) = 2 - 4i + 3i - 6i^2 =2i+6=8i= 2 - i + 6 = 8 - i

Answer: 8i8 - 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=34i\overline{3 + 4i} = -3 - 4i

Correct: 3+4i=34i\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.