Division (Multiplicative Inverse)

Multiply numerator and denominator by the conjugate of the denominator (c − id). This makes the denominator real: (c + id)(c − id) = c² + d². Valid when c + id ≠ 0, i.e., c and d are not both zero.

Division of complex numbers reduces to multiplication once you know how to make the denominator real. The conjugate is the tool.

The Core Idea

We want $\dfrac{a + ib}{c + id}$. The denominator is complex — we cannot directly separate real and imaginary parts. But we know that $(c + id)(c - id) = c^2 + d^2$, which is real and positive (when $c + id \neq 0$).

Multiply numerator and denominator by $c - id$ (the conjugate of the denominator):

$$\frac{a + ib}{c + id} = \frac{(a + ib)(c - id)}{(c + id)(c - id)}$$

Computing the Denominator

$$(c + id)(c - id) = c^2 - (id)^2 = c^2 - i^2 d^2 = c^2 + d^2$$

This is real and equals $|c + id|^2$.

Computing the Numerator

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

The Result

$$\frac{a + ib}{c + id} = \frac{ac + bd}{c^2 + d^2} + i\,\frac{bc - ad}{c^2 + d^2}$$

Both parts are now real numbers, as required.

Multiplicative Inverse

Setting $a = 1$, $b = 0$ gives the inverse of $c + id$:

$$\frac{1}{c + id} = \frac{c}{c^2 + d^2} - i\frac{d}{c^2 + d^2} = \frac{c - id}{c^2 + d^2} = \frac{\bar{z}}{|z|^2}$$

In Euler form: $z = re^{i\theta} \implies z^{-1} = \frac{1}{r}e^{-i\theta}$. Inversion reciprocates the modulus and negates the argument.

Why Multiplying by the Conjugate Works

The identity $z\bar{z} = |z|^2$ is what makes this work. The conjugate is precisely the number that, when multiplied with $z$, removes the imaginary part from the product. This is not a trick — it is the definition of conjugate doing exactly what it was designed to do.