Solving Equations with Complex Roots (Edexcel International A Level Further Maths)

Revision Note

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on 5 April 2024

Quadratics with Complex Roots

How do I solve a quadratic with complex roots?

Quadratic equations with complex roots can still be solved using the quadratic formula or by completing the square
- They have the form $a z^{2} + b z + c = 0$ where $a$ , $b$ and $c$ are real numbers
- For complex roots, the discriminant is negative
  - $b^{2} - 4 a c < 0$
- You must therefore rewrite the square root using $i$
  - $\sqrt{- 16} = 4 i$ , $\sqrt{- 2} = i \sqrt{2}$ , ... and so on
The two roots, $z_{1}$ and $z_{2}$ are complex conjugates of each other
- (provided $a$ , $b$ and $c$ from the equation are all real numbers)
- So $z_{1} = x + i y$ and $z_{2} = x - i y$
  - $z_{2} = z_{1} *$

Examiner Tips and Tricks

Complex conjugate roots is a key concept that is used in many exam questions.

How do I solve a quadratic with one known complex root?

If you know one complex root, the other will be its complex conjugate
- If $z_{1} = 4 + 5 i$ is a root to $z^{2} - 8 z + 41 = 0$
  - then $z_{2} = 4 - 5 i$ must be the other root
  - No solving is required!

How do I find a quadratic from its complex root?

If you are given one complex root, $z_{1}$ , the other root is its complex conjugate, $z_{1} *$
You can then write its quadratic equation in factorised form
- $(z - z_{1}) (z - z_{1} *) = 0$
- Expand this to give the equation in expanded form $a z^{2} + b z + c = 0$
  - Any multiple of this equation also works
A good trick when expanding is to group the first two terms in each bracket
- So $(z - (2 + 3 i)) (z - (2 - 3 i))$ becomes $((z - 2) - 3 i) ((z - 2) + 3 i)$
- Then use the difference of two squares
  - ${(z - 2)}^{2} - {(3 i)}^{2} \equiv {(z - 2)}^{2} + 9 \equiv z^{2} - 4 z + 4 + 9$
- The equation is therefore $z^{2} - 4 z + 13 = 0$

Examiner Tips and Tricks

Always check how the question wants the answer, for example asking for integer coefficients.

Worked Example

(a) Solve $z^{2} - 2 z + 37 = 0$ .

Use the quadratic formula with $a = 1$ , $b = - 2$ and $c = 37$
Find the discriminant $b^{2} - 4 a c$

$\begin{array}{rcl} b^{2} - 4 a c & = & {(- 2)}^{2} - 4 \times 1 \times 37 \\ = & 4 - 148 \\ = & - 144 \end{array}$

Substitute these values into the quadratic formula $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$\begin{array}{rcl} z & = & \frac{- (- 2) \pm \sqrt{- 144}}{2 \times 1} \\ = & \frac{2 \pm \sqrt{- 144}}{2} \end{array}$

Use that $\sqrt{- 144} = \sqrt{144 i^{2}} = 12 i$

$z = \frac{2 \pm 12 i}{2}$

Simplify the answers
Check that they are complex conjugates of each other

$z = 1 \pm 6 i$

(b) If $- \frac{1}{2} + \frac{3 i}{2}$ is one root of the equation $a z^{2} + b z + c = 0$ where $a$ , $b$ and $c$ are positive integers, find $a$ , $b$ and $c$ .

Write down the other root (the complex conjugate)

$- \frac{1}{2} - \frac{3 i}{2}$

Write down the quadratic equation in factorised form, $(z - z_{1}) (z - z_{1} *) = 0$

$(z - (- \frac{1}{2} + \frac{3 i}{2})) (z - (- \frac{1}{2} - \frac{3 i}{2})) = 0$

Expand inside each bracket

$(z + \frac{1}{2} - \frac{3 i}{2}) (z + \frac{1}{2} + \frac{3 i}{2}) = 0$

Group the first two terms in each bracket

$((z + \frac{1}{2}) - \frac{3 i}{2}) ((z + \frac{1}{2}) + \frac{3 i}{2}) = 0$

Use the difference of two squares to expand

${(z + \frac{1}{2})}^{2} - {(\frac{3 i}{2})}^{2} = 0$

Expand and collect, using $i^{2} = - 1$

$\begin{array}{rcl} z^{2} + z + \frac{1}{4} + \frac{9}{4} & = & 0 \\ z^{2} + z + \frac{5}{2} & = & 0 \end{array}$

The question wants the final answer to have positive integer coefficients
Multiply both sides of the equation by 2

$2 z^{2} + 2 z + 5 = 0$

Write down the values of $a$ , $b$ and $c$

$a = 2$ , $b = 2$ and $c = 5$

Positive integer multiples of these answers are also accepted

Cubics & Quartics with Complex Roots

How do I solve a cubic with complex roots?

Cubic equations have either
- 3 real roots
- or 1 real root and one complex conjugate pair of roots
Solve $a z^{3} + b z^{2} + c z + d = 0$ , given that $m + n i$ is a root
- Another root is $m - n i$
  - Its complex conjugate
- So $(z - (m + n i))$ and $(z - (m - n i))$ are factors
- Multiply the factors together to get a quadratic factor
  - $(A z^{2} + B z + C)$
- You need to find the remaining linear factor $(P z + Q)$
- Either use equating coefficients to find $α$
  - $(A z^{2} + B z + C) (P z + Q) \equiv a x^{3} + b x^{2} + c x + d$
  - Expand and simplify the left-hand side
  - Match each coeffi