Further Maths/Lessons/Complex Numbers/Multiplying & Dividing Complex Numbers
Core Pure · Complex Numbers · Year 1

Multiplying & Dividing Complex Numbers

Multiplication expands like any bracket; division needs one clever trick — the conjugate — to clear i from the denominator entirely.

15 min read 3 worked examples

Multiplying complex numbers

Method 2.1 — multiplication

Multiply two complex numbers exactly as you would expand two brackets, then use $i^2=-1$ to simplify:

$$(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i$$

Example 1 · Multiplying two complex numbers
Simplify $(3+2i)(4-5i)$.

$(3+2i)(4-5i)=12-15i+8i-10i^2=12-15i+8i-10(-1)=12+10-7i=22-7i$.

The complex conjugate

Definition 2.2 — complex conjugate

For $z=a+bi$, the complex conjugate is $z^*=a-bi$ (also written $\bar z$). Geometrically, conjugation reflects a point in the real axis. Conjugates satisfy: $z+z^*=2a=2\operatorname{Re}(z)$ (always real), and $zz^*=a^2+b^2$ (always real and non-negative).

Proofzz* is always real and non-negative
1
Let $z=a+bi$, so $z^*=a-bi$.
Definition of the conjugate.
2
$zz^*=(a+bi)(a-bi)=a^2-abi+abi-b^2i^2$
Expand the product directly.
3
$=a^2-b^2(-1)=a^2+b^2$
Using $i^2=-1$; the cross terms cancel exactly.
4
Since $a,b\in\mathbb{R}$, $a^2\geq0$ and $b^2\geq0$, so $a^2+b^2\geq0$ and is real.
Sum of two real squares.
∎

Dividing complex numbers

Method 2.3 — division by realising the denominator

To simplify $\dfrac{z_1}{z_2}$, multiply numerator and denominator by the conjugate of the denominator, $z_2^*$. Since $z_2z_2^*$ is real, this eliminates $i$ from the denominator entirely.

Example 2 · Dividing complex numbers
Express $\dfrac{5+i}{2-3i}$ in the form $a+bi$.

Multiply top and bottom by $2+3i$ (the conjugate of $2-3i$):

$$\frac{5+i}{2-3i}\times\frac{2+3i}{2+3i}=\frac{(5+i)(2+3i)}{(2-3i)(2+3i)}=\frac{10+15i+2i+3i^2}{4+9}=\frac{10+17i-3}{13}=\frac{7+17i}{13}$$

So $\dfrac{5+i}{2-3i}=\dfrac{7}{13}+\dfrac{17}{13}i$.

Example 3 · Combined operations
Given $z=3-4i$, find $z+\dfrac{20}{z}$ in the form $a+bi$.

$\dfrac{20}{z}=\dfrac{20}{3-4i}\times\dfrac{3+4i}{3+4i}=\dfrac{20(3+4i)}{9+16}=\dfrac{60+80i}{25}=\dfrac{12}{5}+\dfrac{16}{5}i$.

$z+\dfrac{20}{z}=(3-4i)+(\tfrac{12}{5}+\tfrac{16}{5}i)=(3+\tfrac{12}{5})+(-4+\tfrac{16}{5})i=\tfrac{27}{5}-\tfrac{4}{5}i$.

Common mistake

When multiplying $(a+bi)(a-bi)$, students sometimes forget the middle terms cancel and mistakenly write $a^2-b^2$ (as if factorising a difference of squares over the reals). Because of the extra factor of $i^2=-1$, the correct result is $a^2+b^2$ — always a sum, never a difference.

Exam technique

Whenever a question asks you to "express in the form $a+bi$" and a fraction with a complex denominator appears, realising the denominator by its conjugate is almost always the required first step — examiners look specifically for the line "multiplying numerator and denominator by the conjugate" or equivalent working.

Exam-style questions

Question 1 6 marks

Given $z_1=2-3i$ and $z_2=-1+4i$, find, in the form $a+bi$: (a) $z_1z_2$, (b) $z_1^2$.

(a) $z_1z_2=(2-3i)(-1+4i)=-2+8i+3i-12i^2=-2+11i+12=10+11i$ (M1, A1, A1)

(b) $z_1^2=(2-3i)^2=4-12i+9i^2=4-12i-9=-5-12i$ (M1, A1, A1)

Question 2 6 marks

Express $\dfrac{7-4i}{1+2i}$ in the form $a+bi$, and hence find $\left|\dfrac{7-4i}{1+2i}\right|^2$ using your answer (i.e. compute $a^2+b^2$).

$\dfrac{7-4i}{1+2i}\times\dfrac{1-2i}{1-2i}=\dfrac{(7-4i)(1-2i)}{1+4}=\dfrac{7-14i-4i+8i^2}{5}=\dfrac{7-18i-8}{5}=\dfrac{-1-18i}{5}$ (M1, A1, A1)

$=-\tfrac15-\tfrac{18}{5}i$ (A1)

$a^2+b^2=\left(\tfrac15\right)^2+\left(\tfrac{18}{5}\right)^2=\tfrac{1}{25}+\tfrac{324}{25}=\tfrac{325}{25}=13$ (M1, A1)

Question 3 6 marks

Given that $z=x+yi$ satisfies $z+2z^*=9+5i$, find the values of $x$ and $y$.

$z+2z^*=(x+yi)+2(x-yi)=3x-yi$ (M1, A1)

Equating: $3x=9 \Rightarrow x=3$; $-y=5 \Rightarrow y=-5$ (M1, A1, A1, A1)

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