Complex arithmetic: add, subtract, multiply and divide

9 minA Level3.9.2

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can add, subtract, multiply and divide complex numbers written as $x + iy$ , and leave each answer in that same form with no $i$ hiding in a denominator.

Key to the symbols

Symbol	Meaning
$a,\ b,\ c,\ d$	ordinary real numbers, used as placeholders
$i$	the number defined by $i^2 = -1$
$z^*$	the conjugate of $z$ (defined below)
$\operatorname{Re} z$	the real part of $z$
$\operatorname{Im} z$	the imaginary part of $z$ , which is a real number

A complex number has two separate parts, so an answer is only finished when both parts are shown separately.

What "Cartesian form" means

Cartesian form (also called rectangular form) means writing a complex number as

$x + iy$

where $x$ and $y$ are real numbers.

$x$ is the real part, $\operatorname{Re} z$ .
$y$ is the imaginary part, $\operatorname{Im} z$ . It is the number in front of $i$ , so it does not include the $i$ .

For example, $4 - 3i$ is in Cartesian form with $x = 4$ and $y = -3$ .

Cambridge expects a final answer in this shape. Something like $\dfrac{2}{1+i}$ is not finished, because the $i$ is still in the denominator. The whole point of the division method in this lesson is to remove it.

Adding and subtracting: collect like terms

Treat $i$ as a letter, like $x$ , and collect terms of the same kind.

$(a + ib) + (c + id) = (a + c) + i(b + d)$

$(a + ib) - (c + id) = (a - c) + i(b - d)$

Real parts combine with real parts. Imaginary parts combine with imaginary parts. The two kinds never mix.

For subtraction, the minus sign applies to both parts of the second bracket. This is the same rule as $5 - (2 + 3)$ being $5 - 2 - 3$ , not $5 - 2 + 3$ .

Worked example: A sum and a difference

Simplify $(4 + 2i) + (1 - 5i) - (2 - i)$ .

Have a go first?

Why the two parts stay separate

On an Argand diagram, the real part is measured across and the imaginary part is measured up.

Adding complex numbers moves you across by the sum of the real parts and up by the sum of the imaginary parts.
A real number cannot create any movement up, and a multiple of $i$ cannot create any movement across.

That is why one complex equation, like $a + bi = c + di$ , gives you two ordinary equations, $a = c$ and $b = d$ .

Multiplying: expand, then use i² = −1

Multiply out the brackets exactly as you would with $(a + bx)(c + dx)$ , then replace every $i^2$ with $-1$ .

$(a + ib)(c + id) = ac + iad + ibc + i^2bd$

The last term contains $i^2$ , so it turns into $-bd$ :

$= (ac - bd) + i(ad + bc)$

Two warnings.

Do not multiply the two "i" terms and leave $i^2$ in the answer. Replace it immediately.
Do not think the real part of the answer is just $ac$ . The cross terms $ad$ and $bc$ join the imaginary part, and $bd$ turns the real part into $ac - bd$ .

Worked example: Multiply (3 + 2i)(1 − 4i)

Find the product, showing full working.

Have a go first?

Worked example: Squaring a complex number

Find $(2 - i)^2$ .

Have a go first?

Check addition and multiplication

Short answerQuestion 1 · Recall

Answer the 2 questions above to continue.