Inverse of a complex number

Statement

Let $z$ be a complex number with $z \neq 0$. Then $$z^{-1} = \frac{\bar{z}}{|z|^2}.$$ With $z = a + b i$ and $a, b$ real, this is $z^{-1} = \dfrac{a}{a^2 + b^2} - \dfrac{b}{a^2 + b^2} \, i$.

Remarks

Proof: $z \bar{z} = |z|^2$ by the formula for a complex number times its conjugate, and $|z|^2 > 0$ because $z \neq 0$. Divide by $|z|^2$. To compute a quotient $w / z$, multiply the numerator and the denominator by $\bar{z}$.

Explanations

L1 human

To divide by a complex number, multiply above and below by its conjugate. Then the denominator becomes a positive real number. For example, $\dfrac{1}{1 + i} = \dfrac{1 - i}{(1 + i)(1 - i)} = \dfrac{1 - i}{2}$.