Solving Equations with Complex Roots (Edexcel IAL Further Maths)

Quadratics with Complex Roots

Spec point — 1.5

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 $az^{2}+bz+c=0$ where $a$, $b$ and $c$ are real numbers

  • For complex roots, the discriminant is negative

    • $b^{2}-4ac<0$

  • You must therefore rewrite the square root using $i$

    • $\sqrt{-16}=4i$, $\sqrt{-2}=i\sqrt{2}$, ... 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 $z_{1}=x+iy$ and $z_{2}=x-iy$

    • $z_{2}=z_{1}\ast$

Exam Hint 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 $z_{1}=4+5i$ is a root to $z^{2}-8z+41=0$

    • then $z_{2}=4-5i$ 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, $z_{1}\ast$

• You can then write its quadratic equation in factorised form

  • $( z - z_{1} ) ( z - z_{1} \text{asterisk} \times ) = 0$

  • Expand this to give the equation in expanded form $az^{2}+bz+c=0$

    • Any multiple of this equation also works

• A good trick when expanding is to group the first two terms in each bracket

  • So $( \mathrm{z} - ( 2 + 3 \mathrm{i} ) ) ( \mathrm{z} - ( 2 - 3 \mathrm{i} ) )$ becomes $( ( z - 2 ) - 3 \mathrm{i} ) ( ( z - 2 ) + 3 \mathrm{i} )$

  • Then use the difference of two squares

    • $( z - 2 )^{2} - ( 3 \mathrm{i} )^{2} \equiv ( z - 2 )^{2} + 9 \equiv z^{2} - 4 z + 4 + 9$

  • The equation is therefore $z^{2}-4z+13=0$

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

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

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

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

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

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

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

$z=\frac{2\pm12i}{2}$

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

$z=1\pm6i$

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

Write down the other root (the complex conjugate)

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

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

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

Expand inside each bracket

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

Group the first two terms in each bracket

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

Use the difference of two squares to expand

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

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

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

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

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

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

Final answer: $a=2$,$b=2$and$c=5$

Positive integer multiples of these answers are also accepted

Cubics & Quartics with Complex Roots

Spec point — 1.6

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 $az^{3}+bz^{2}+cz+d=0$, given that $m+ni$ is a root

  • Another root is $m-ni$

    • Its complex conjugate

  • So $( z - ( m + n \mathrm{i} ) )$ and $(z-(m-ni))$ are factors

  • Multiply the factors together to get a quadratic factor

    • $(Az^{2}+Bz+C)$

  • You need to find the remaining linear factor $(Pz+Q)$

  • 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 $az^{4}+bz^{3}+cz^{2}+dz+e=0$, given that $m+ni$ is a root

  • Another root is $m-ni$

    • Its complex conjugate

  • So $( z - ( m + n \mathrm{i} ) )$ and $(z-(m-ni))$ are factors

  • Multiply the factors together to get a quadratic factor

    • $(Az^{2}+Bz+C)$

  • 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 $Pz^{2}+Qz+R=0$

  • 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 $( z - \\alpha )$

  • Complex roots give quadratic factors (from conjugate pairs)

  • $z=m+ni$ gives $( z - ( m + n \mathrm{i} ) ) ( z - ( m - n \mathrm{i} ) ) \equiv A z^{2} + B z + C$

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

Write down the other complex root by finding the complex conjugate

$-4-2i$

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

$( z - ( - 4 + 2 \mathrm{i} ) ) ( z - ( - 4 - 2 \mathrm{i} ) )$

Expand inside each bracket

$= ( z + 4 - 2 \mathrm{i} ) ( z + 4 + 2 \mathrm{i} )$

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

$\begin{gathered} = ( ( z + 4 ) - 2 \mathrm{i} ) ( ( z + 4 ) + 2 \mathrm{i} ) \\ = ( z + 4 )^{2} - ( 2 \mathrm{i} )^{2} \\ = z^{2} + 8 z + 16 + 4 \\ = z^{2} + 8 z + 20 \end{gathered}$

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

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

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

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

$z^{3}+5z^{2}-4z-60=0$ 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

Final answer: $z=-4+2i$,$-4-2i$or$3$

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

Write down the other complex root by finding the complex conjugate

$1-3i$

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

$( z - ( 1 + 3 \mathrm{i} ) ) ( z - ( 1 - 3 \mathrm{i} ) )$

Expand inside each bracket

$= ( z - 1 - 3 \mathrm{i} ) ( z - 1 + 3 \mathrm{i} )$

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

$\begin{gathered} = ( ( z - 1 ) - 3 \mathrm{i} ) ( ( z - 1 ) + 3 \mathrm{i} ) \\ = ( z - 1 )^{2} - ( 3 \mathrm{i} )^{2} \\ = z^{2} - 2 z + 1 + 9 \\ = z^{2} - 2 z + 10 \end{gathered}$

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

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

To get 90 on the left, let $Q^{2}=9$ so $Q=\pm3$ To find out which value $Q$ takes, expand and collect the right-hand side further

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

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

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

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

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

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

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

Write out the answers clearly

Final answer: $A=31$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