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

Revision Note

Author

Mark Curtis

Expertise

Maths

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 $z_{2} = x - i y$
  - $z_{2} = z_{1} *$

Exam Tip

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 $z^{2} - 4 z + 13 = 0$

Exam Tip

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 coefficient with the right-hand side
- Or use polynomial division to find $(P z + Q)$
  - $(a z^{3} + b z^{2} + c z + d) \div (A z^{2} + B z + C)$
- Write out your three roots clearly

How do I solve a quartic with complex roots?

Quartic equations have either
- 4 real roots
- or 2 real roots and 1 pair of complex conjugate roots
- or 2 pairs of complex conjugate roots
Solve $a z^{4} + b z^{3} + c z^{2} + d z + e = 0$ , 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 quadratic factor $(P z^{2} + Q z + R)$
- Either use equating coefficients to find $P$ , $Q$ and $R$
  - $(A z^{2} + B z + C) (P z^{2} + Q z + R) \equiv a z^{4} + b z^{3} + c z^{2} + d z + e$
  - Expand and simplify the left-hand side
  - Match each coefficient with the right-hand side
- Or use polynomial division to find $(P z^{2} + Q z + R)$
  - $(a z^{4} + b z^{3} + c z^{2} + d z + e) \div (A z^{2} + B z + C)$
- Solve $P z^{2} + Q z + R = 0$
- Write out your four roots clearly

How do I find unknown coefficients of equations?

Find as many factors as possible from given roots
- Real roots give linear factors
  - $z = α$ gives $(z - α)$
- Complex roots give quadratic factors (from conjugate pairs)
- $z = m + n i$ gives $(z - (m + n i)) (z - (m - n i)) \equiv A z^{2} + B z + C$
Then write out the rest algebraically and use equating coefficients
- Start by matching coefficients of the highest powers and of the constants
  - These can often been seen by inspection

Exam Tip

Whilst polynomial division can be used instead of equating coefficients, it is not recommended when exam questions have algebraic coefficients!

Worked Example

(a) Given that $- 4 + 2 i$ is a root of the equation $z^{3} + 5 z^{2} - 4 z - 60 = 0$ , find the other two roots.

Write down the other complex root by finding the complex conjugate

$- 4 - 2 i$

Write down a factorised quadratic factor of the cubic
Use the form $(z - (- 4 + 2 i)) (z - (- 4 - 2 i))$

$(z - (- 4 + 2 i)) (z - (- 4 - 2 i))$

Expand inside each bracket

$= (z + 4 - 2 i) (z + 4 + 2 i)$

Group the first pair of terms in each bracket
Then expand using the difference between two squares
Simplify the quadratic factor

$\begin{array}{rcl} = & ((z + 4) - 2 i) ((z + 4) + 2 i) \\ = & {(z + 4)}^{2} - {(2 i)}^{2} \\ = & z^{2} + 8 z + 16 + 4 \\ = & z^{2} + 8 z + 20 \end{array}$

Write the cubic as the quadratic factor multiplied by a linear factor $P z + Q$

$z^{3} + 5 z^{2} - 4 z - 60 \equiv (z^{2} + 8 z + 20) (P z + Q)$

Either expand the right-hand side and equate coefficients
Or see that $P = 1$ to get $z^{3}$ on the left, and $Q = - 3$ to get $- 60$ on the left

$z^{3} + 5 z^{2} - 4 z - 60 \equiv (z^{2} + 8 z + 20) (z - 3)$

$z^{3} + 5 z^{2} - 4 z - 60 = 0$ so set the factorised cubic equal to zero

$(z^{2} + 8 z + 20) (z - 3) = 0$

The solutions from the quadratic factor are the complex conjugates
The solution from $(z - 3) = 0$ is $z = 3$
Write down all three solutions

$z = - 4 + 2 i$ , $- 4 - 2 i$ or $3$

(b) It is known that the equation $z^{4} - 8 z^{3} + A z^{2} + B z + 90$ has a complex root $1 + 3 i$ and a repeated real root. Find $A$ and $B$ .

Write down the other complex root by finding the complex conjugate

$1 - 3 i$

Write down a factorised quadratic factor of the quartic
Use the form $(z - (1 + 3 i)) (z - (1 - 3 i))$

$(z - (1 + 3 i)) (z - (1 - 3 i))$

Expand inside each bracket

$= (z - 1 - 3 i) (z - 1 + 3 i)$

Group the first pair of terms in each bracket
Then expand using the difference between two squares
Simplify the quadratic factor

$\begin{array}{rcl} = & ((z - 1) - 3 i) ((z - 1) + 3 i) \\ = & {(z - 1)}^{2} - {(3 i)}^{2} \\ = & z^{2} - 2 z + 1 + 9 \\ = & z^{2} - 2 z + 10 \end{array}$

The other two roots are repeated and real
Write the quartic as the quadratic factor multiplied by the square of a linear factor ${(P z + Q)}^{2}$

$z^{4} - 8 z^{3} + A z^{2} + B z + 90 \equiv (z^{2} - 2 z + 10) {(P z + Q)}^{2}$

Imagine expanding the right-hand side
To get $z^{4}$ on the left, you can let $P$ equal 1

$\begin{array}{rcl} z^{4} - 8 z^{3} + A z^{2} + B z + 90 & \equiv & (z^{2} - 2 z + 10) {(z + Q)}^{2} \\ \equiv & (z^{2} - 2 z + 10) (z^{2} + 2 Q z + Q^{2}) \end{array}$

To get 90 on the left, let $Q^{2} = 9$ so $Q = \pm 3$
To find out which value $Q$ takes, expand and collect the right-hand side further

$\begin{array}{rcl} z^{4} - 8 z^{3} + A z^{2} + B z + 90 & \equiv & z^{4} + 2 Q z^{3} + 9 z^{2} - 2 z^{3} - 4 Q z^{2} - 18 z + 10 z^{2} + 20 Q z + 90 \\ \equiv & z^{4} + (2 Q - 2) z^{3} + (9 - 4 Q + 10) z^{2} + (- 18 + 20 Q) z + 90 \end{array}$

The $z^{4}$ and $+ 90$ are correct
Equate the $z^{3}$ coefficients to get an equation in $Q$ and solve

$\begin{array}{rcl} - 8 & = & 2 Q - 2 \\ - 6 & = & 2 Q \\ - 3 & = & Q \end{array}$

Substitute $Q = - 3$ into the coefficient of $z^{2}$ to get $A$

$\begin{array}{rcl} A & = & 9 - 4 (- 3) + 10 \\ = & 31 \end{array}$

Substitute $Q = - 3$ into the coefficient of $z$ to get $B$

$\begin{array}{rcl} B & = & - 18 + 20 (- 3) \\ = & - 78 \end{array}$

Write out the answers clearly

$A = 31$ and $B = - 78$

Polynomial division is not recommended as $A$ and $B$ are algebraic
$P = - 1$ and $Q = 3$ also work, as the bracket ${(P z + Q)}^{2}$ is squared

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

Downloadable PDFs
Unlimited Revision Notes
Topic Questions
Past Papers
Model Answers
Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Next topic

Did this page help you?

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

Revision Note

Quadratics with Complex Roots

How do I solve a quadratic with complex roots?

Exam Tip

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

How do I find a quadratic from its complex root?

Exam Tip

Worked Example

Cubics & Quartics with Complex Roots

How do I solve a cubic with complex roots?

How do I solve a quartic with complex roots?

How do I find unknown coefficients of equations?

Exam Tip

Worked Example

You've read 0 of your 0 free revision notes

Get unlimited access

Join the 100,000+ Students that ❤️ Save My Exams

Author: Mark Curtis