Complex Numbers – Study Notes
Program: DCECE
Subject: Mathematics
Topic: Complex Numbers
1. Simple Explanation
A complex number is an extension of the real numbers that allows us to solve equations that have no real solutions, such as (x^2 + 1 = 0).
It is written as
[ z = a + bi ]
- (a) – real part
- (b) – imaginary part
- (i) – imaginary unit, defined by (i^2 = -1)
Complex numbers form a plane (the complex plane or Argand diagram) where the horizontal axis is the real part and the vertical axis is the imaginary part.
2. Key Points
| Concept | Definition / Formula | Notes |
|---|---|---|
| Imaginary Unit | (i^2 = -1) | Fundamental to all complex operations |
| Conjugate | (\overline{z} = a - bi) | Used to rationalise denominators, find modulus |
| Modulus (Absolute Value) | ( | z |
| Argument (Angle) | (\theta = \tan^{-1}!\left(\frac{b}{a}\right)) | Angle with positive real axis (use atan2 for correct quadrant) |
| Polar Form | (z = r(\cos\theta + i\sin\theta) = re^{i\theta}) | (r = |
| Addition/Subtraction | ((a+bi)+(c+di) = (a+c)+(b+d)i) | Combine like terms |
| Multiplication | ((a+bi)(c+di) = (ac - bd) + (ad + bc)i) | Apply distributive law, use (i^2=-1) |
| Division | (\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{c^2+d^2}) | Multiply numerator & denominator by conjugate |
| Euler’s Formula | (e^{i\theta} = \cos\theta + i\sin\theta) | Links complex exponentials to trig functions |
| De Moivre’s Theorem | ((\cos\theta + i\sin\theta)^n = \cos(n\theta)+i\sin(n\theta)) | Useful for powers and roots |
| Roots of Unity | (e^{2\pi i k/n}) for (k=0,\dots,n-1) | Solutions to (z^n = 1) |
3. Worked Example
Problem: Find the product and quotient of
[
z_1 = 3 + 4i , \quad z_2 = 1 - 2i
]
3.1 Multiplication
[ \begin{aligned} z_1 z_2 &= (3+4i)(1-2i) \ &= 3\cdot1 + 3(-2i) + 4i\cdot1 + 4i(-2i) \ &= 3 - 6i + 4i - 8i^2 \ &= 3 - 2i - 8(-1) \quad (\text{since } i^2=-1) \ &= 3 - 2i + 8 \ &= 11 - 2i \end{aligned} ]
Result: (z_1 z_2 = 11 - 2i)
3.2 Division
[ \frac{z_1}{z_2} = \frac{3+4i}{1-2i} ]
Multiply numerator & denominator by conjugate of denominator ((1+2i)):
[ \begin{aligned} \frac{3+4i}{1-2i} &= \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)} \ &= \frac{(3\cdot1 + 3\cdot2i + 4i\cdot1 + 4i\cdot2i)}{1^2 - (2i)^2} \ &= \frac{3 + 6i + 4i + 8i^2}{1 - 4i^2} \ &= \frac{3 + 10i + 8(-1)}{1 - 4(-1)} \ &= \frac{3 + 10i - 8}{1 + 4} \ &= \frac{-5 + 10i}{5} \ &= -1 + 2i \end{aligned} ]
Result: (\displaystyle \frac{z_1}{z_2} = -1 + 2i)
4. Quick Revision Summary
- Complex number: (z = a + bi)
- Conjugate: (\overline{z} = a - bi)
- Modulus: (|z| = \sqrt{a^2 + b^2})
- Argument: (\theta = \tan^{-1}!\left(\frac{b}{a}\right)) (use
atan2for correct quadrant) - Polar form: (z = r(\cos\theta + i\sin\theta) = re^{i\theta})
- Addition/Subtraction: combine like terms
- Multiplication: ((a+bi)(c+di) = (ac-bd)+(ad+bc)i)
- Division: multiply by conjug