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

Revision Note

Written by: Mark Curtis

Reviewed by: Dan Finlay

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 $i^{2} = - 1$ then collect the real parts

$\begin{array}{rcl} = & 8 + 38 i + 9 \times (- 1) \\ = & 8 + 38 i - 9 \end{array}$

$- 1 + 38 i$

(b) ${(3 - 4 i)}^{2}$

Write using double brackets then expand

$= (3 - 4 i) (3 - 4 i) = 9 - 12 i - 12 i + 16 i^{2}$

Collect the imaginary parts
Use that $i^{2} = - 1$ then collect the real parts

$\begin{array}{rcl} = & 9 - 24 i + 16 \times (- 1) \\ = & 9 - 24 i - 16 \end{array}$

$\begin{array}{rcl} - 7 - 24 i \end{array}$

Use index laws to move the power onto the individual terms

${(- 2)}^{11} \times i^{11}$

Work out ${(- 2)}^{11}$

${(- 2)}^{11} = - 2048$

Work out $i^{11}$
It helps to write it in terms of $i^{10}$ then $i^{2}$
Use $i^{2} = - 1$

$\begin{array}{rcl} i^{11} & = & i^{10} \times i \\ = & {(i^{2})}^{5} \times i \\ = & {(- 1)}^{5} \times i \\ = & - 1 \times i \\ = & - i \end{array}$

Multiply both parts together
Two minus signs make a plus

$- 2048 \times (- i)$

$2048 i$

Complex Conjugates & Division

What is a complex conjugate?

If $z = x + y i$ then the complex conjugate of $z$ is $z * = x - y i$
- The sign of the imaginary part changes
Note that
- $z + z *$ is always real
  - since $x + y i + x - y i = 2 x$
- $z - z *$ is always imaginary
  - since $x + y i - (x - y i) = 2 y i$
- $z z *$ is always real (and non-negative)
  - since $(x + y i) (x - y i) = x^{2} - y^{2} i^{2} = x^{2} + y^{2}$
  - $z z * = z * z$

How do I divide complex numbers?

To divide $z_{1}$ by $z_{2}$ , multiply top and bottom of $\frac{z_{1}}{z_{2}}$ by $z_{2} *$
- $z_{2} *$ is the complex conjugate of the denominator
This makes the denominator a real number
- which allows you to write the final answer in Cartesian form, $x + y i$
The process is called realising the denominator
- It is a very similar to rationalising the denominator with surds
For example, to work out $\frac{50 + 75 i}{3 + 4 i}$
- calculate $\frac{(50 + 75 i)}{(3 + 4 i)} \times \frac{(3 - 4 i)}{(3 - 4 i)}$
  - It helps to write the brackets in
- then expand and simplify

Examiner Tips and Tricks

To check your answer in an exam, multiply it by the denominator and see if you get the numerator.

Worked Example

Let $z_{1} = 1 + 7 i$ and $z_{2} = 3 - i$ .

Find and simplify $\frac{z_{1}}{z_{2}}$ , giving your answer in the form $x + y i$ where $x$ and $y$ are real numbers.

Find the complex conjugate of the denominator

$z_{2} * = 3 + i$

Multiply the top and bottom of $\frac{z_{1}}{z_{2}}$ by $z_{2} *$

$\frac{(1 + 7 i)}{(3 - i)} \times \frac{(3 + i)}{(3 + i)}$

Write as one single fraction then expand top and bottom separately

$\frac{(1 + 7 i) (3 + i)}{(3 - i) (3 + i)} = \frac{3 + i + 21 i + 7 i^{2}}{9 - i^{2}}$

Use that $i^{2} = - 1$ then collect real and imaginary parts

$\begin{array}{rcl} = & \frac{3 + i + 21 i - 7}{9 - (- 1)} \\ = & \frac{- 4 + 22 i}{10} \end{array}$

To give your answer in the form $x + y i$ , split the fraction then simplify

$- \frac{4}{10} + \frac{22}{10} i$

$- \frac{2}{5} + \frac{11}{5} i$

$- 0.4 + 2.2 i$ is also accepted

Last updated: 12 April 2024

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Next:Modulus & Argument

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

Revision Note

Cartesian Form of Complex Numbers

What is an imaginary number?

What is a complex number?

How are real numbers and complex numbers related?

How do I add or subtract complex numbers?

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

Multiplying Complex Numbers

How do I multiply complex numbers?

How do I find powers of i?

Complex Conjugates & Division

What is a complex conjugate?

How do I divide complex numbers?

You've read 0 of your 5 free revision notes this week

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Complex Numbers

Operations with Complex Numbers

Introduction to Complex Numbers

Modulus & Argument

Equating Real & Imaginary Parts

Solving Equations with Complex Roots

Solving Equations with Complex Roots

Roots of Quadratic Equations

Roots of Quadratic Equations

Roots of Quadratic Equations

Forming Quadratics with New Roots

Numerical Solutions of Equations

Numerical Solutions of Equations

Roots in Intervals

Interval Bisection

Linear Interpolation

Newton-Raphson Method

Coordinate Systems

Coordinate Systems

The Parabola

The Rectangular Hyperbola

Tangents & Normals

Matrices

Matrix Algebra

Introduction to Matrices

Multiplying Matrices

Inverse Matrices

Proving Matrix Relationships

Transformations using Matrices

Transformations using Matrices

Standard Matrix Transformations

Inverse Matrix Transformations

Combinations of Matrix Transformations

Series

Series

Standard Series

Combinations of Series

Proof

Proof by Induction

Proof by Induction

Standard Proofs by Induction

Author: Mark Curtis

Author: Dan Finlay