Introduction to Complex Numbers (Edexcel International A Level Further Maths)

Revision Note

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on 12 April 2024

Cartesian Form of Complex Numbers

What is an imaginary number?

Equations like $x^{2} = - 9$ have no real solutions
- Squaring real numbers always gives a positive value
  - No real number squared could give -9
- $x = \pm 3$ are real numbers, but neither work
  - They give +9
To get round this, mathematicians introduce the imaginary number, $i$ , as follows:
- $i^{2} = - 1$
  - This can be thought of as $i = \sqrt{- 1}$
Rules for surds and indices can be used
- $x^{2} = - 9$ means $x^{2} = 9 \times (- 1) = 9 i^{2}$
  - The imaginary solutions are $x = \pm 3 i$

What is a complex number?

Complex numbers have both a real part and an imaginary part
- For example, $3 + 4 i$
  - The real part is 3
  - The imaginary part is 4
This is called Cartesian form
In general, Cartesian form is written using the notation:
- $z = x + y i$
- $Re (z) = x$
- $Im (z) = y$

The letter $ℂ$ stands for all complex numbers
- $z \in ℂ$
Complex numbers with no imaginary parts are real numbers
- The real numbers, $ℝ$ , are a subset of the complex numbers, $ℂ$
  - $ℝ \subset ℂ$
In Cartesian form $z = x + y i \in ℂ$
- $x \in ℝ$
- $y \in ℝ$ as $y$ itself takes real values
  - Multiplying it by $i$ makes it imaginary: $y i$
Complex numbers with no real parts are called imaginary numbers

How do I add or subtract complex numbers?

Add or subtract their real parts and imaginary parts separately
- $\begin{array}{rcl} (3 + 4 i) + (2 + 8 i) & = & (3 + 2) + (4 + 8) i = 5 + 12 i \end{array}$
- $\begin{array}{rcl} (3 + 4 i) - (2 + 8 i) & = & (3 - 2) + (4 - 8) i = 1 - 4 i \end{array}$

How do I multiply or divide complex numbers by real numbers?

Multiply or divide their real parts and imaginary parts separately
- $\begin{array}{rcl} 10 (3 + 4 i) & = & 30 + 40 i \end{array}$
- $\begin{array}{rcl} (3 + 4 i) \div 10 & = & 0.3 + 0.4 i \end{array}$
  - This can also be written $\begin{array}{rcl} \frac{1}{10} (3 + 4 i) & = & \frac{3}{10} + \frac{4}{10} i \end{array}$

Examiner Tips and Tricks

Avoid these handwriting misinterpretations when writing $i$ in the exam:
- $2 \sqrt{5} i$ can look like $2 \sqrt{5 i}$
  - Alternatives are $(2 \sqrt{5}) i$ or $2 i \sqrt{5}$
- $\frac{3}{2} i$ can look like $\frac{3}{2 i}$
  - An alternative is $\frac{3 i}{2}$

Worked Example

Two complex numbers are given by $z_{1} = p + 2 i$ and $z_{2} = - 7 + q i$ , where $p$ and $q$ are real.

Given that $z_{1} + 2 z_{2} = 4 - 8 i$ , find $p$ and $q$ .

Substitute the complex numbers into the left-hand side

$(p + 2 i) + 2 (- 7 + q i)$

Expand the brackets and collect real and imaginary terms

$\begin{array}{rcl} = & p + 2 i - 14 + 2 q i \\ = & (p - 14) + (2 + 2 q) i \end{array}$

Compare this to the right-hand side
Set the real parts equal to each other and solve

$\begin{array}{rcl} p - 14 & = & 4 \\ p & = & 18 \end{array}$

Set the imaginary parts equal to each other and solve

$\begin{array}{rcl} 2 + 2 q & = & - 8 \\ 2 q & = & - 10 \\ q & = & - 5 \end{array}$

$p = 18$ and $q = - 5$

Multiplying Complex Numbers

How do I multiply complex numbers?

All rules of expanding brackets still work
- You need to remember that $i^{2} = - 1$
For example, $(a + b i) (c + d i) = a c + a d i + b c i + b d i^{2}$
- Use $i^{2} = - 1$ in the last term
  - $a c + a d i + b c i - b d$
- Then group real and imaginary parts
  - Factorise out the $i$
  - $a c - b d + (a d + b c) i$
Note that the difference between two squares becomes
- $(a + b i) (a - b i) = a^{2} - b^{2} i^{2} = a^{2} + b^{2}$

How do I find powers of i?

Use the fact that $i^{2} = - 1$
- Below are the first few powers of $i$ :
  - $i^{0} = 1$
  - $i^{1} = i$
  - $i^{2} = - 1$
  - $i^{3} = - i$ from $i^{2} \times i = (- 1) \times i$
  - $i^{4} = 1$ from ${(i^{2})}^{2} = {(- 1)}^{2} = 1$
  - $i^{5} = i$ from $i^{5} = {(i^{2})}^{2} \times i = i$
- The pattern above continues
  - $i^{6} = - 1$
  - $i^{7} = - i$
Find higher powers of $i$ using a base of $i^{2}$
- Remember that -1 to an even power is 1 (or to an odd power is -1)
  - $i^{23} = {(i^{2})}^{11} \times i = {(- 1)}^{11} \times i = - i$

Examiner Tips and Tricks

Questions that say "show your working clearly" won't accept answers written down from a calculator.

Worked Example

Showing your working clearly, find and simplify:

(a) $(4 + i) (2 + 9 i)$

Expand the brackets

$= 4 \times 2 + 4 \times 9 i + i \times 2 + i \times 9 i = 8 + 36 i + 2 i + 9 i^{2}$

Collect the imaginary parts
Use that