Dividing Complex Numbers

An i in the denominator

Dividing by a real number is easy: divide both parts. $\left(6 - 4i\right) / 2 = 3 - 2i$.

Dividing by a complex number is not. $\frac{1}{{2 + i}}$ is a number, but in that form its real part and its imaginary part cannot be read off. The aim is to rewrite it as a + bi, and the way there is to make the denominator real.

Multiply top and bottom by the conjugate

The conjugate of the denominator 2 + i is 2 − i, and $\frac{{2 - i}}{{2 - i}} = 1$. Multiplying by 1 does not change the value:

$\frac{1}{{2 + i}} = \frac{{2 - i}}{{\left(2 + i\right)\left(2 - i\right)}}$.

The denominator is a number times its conjugate, which is always real: $\left(2 + i\right)\left(2 - i\right) = 4 - i^{2} = 4 + 1 = 5$. So $\frac{1}{{2 + i}} = \frac{{2 - i}}{5} = 0.4 - 0.2i$.

The denominator (2 + i)(2 − i). The gold pieces 2i and −2i cancel, and $-i^{2} = +1$, so the denominator is the real number 5.

Check by multiplying back

If $\frac{1}{{2 + i}} = 0.4 - 0.2i$, then (2 + i)(0.4 − 0.2i) should be 1. It is: $0.8 - 0.4i + 0.4i - 0.2i^{2} = 0.8 + 0.2 = 1$.

A full division

Work out $\left(5 + 5i\right) / \left(3 + i\right)$. The conjugate of the denominator is 3 − i, so multiply the top and the bottom by 3 − i.

The bottom: $\left(3 + i\right)\left(3 - i\right) = 9 - i^{2} = 9 + 1 = 10$. The top: $\left(5 + 5i\right)\left(3 - i\right) = 15 - 5i + 15i - 5i^{2} = 15 + 10i + 5 = 20 + 10i$.

So $\left(5 + 5i\right) / \left(3 + i\right) = \frac{{20 + 10i}}{10} = 2 + i$. Now the denominator is real, each part is divided by it. Check: $\left(2 + i\right)\left(3 + i\right) = 6 + 2i + 3i + i^{2} = 5 + 5i$.

The new numerator (5 + 5i)(3 − i). Here the i-terms do not cancel: −5i + 15i = 10i. The gold piece is $-5i^{2} = +5$, so the numerator is 20 + 10i.

The rule

To divide a + bi by c + di, multiply the top and the bottom by c − di, the conjugate of the denominator. The bottom becomes $\left(c + \text{di}\right)\left(c - \text{di}\right) = c^{2} + d^{2}$, which is real, and then each part of the top is divided by it.

Flip the sign of the imaginary part whichever sign it has. To divide 3 + 4i by 1 − 2i, multiply by 1 + 2i. The top is $\left(3 + 4i\right)\left(1 + 2i\right) = 3 + 6i + 4i + 8i^{2} = -5 + 10i$ and the bottom is 1 + 4 = 5, so the answer is −1 + 2i. Check: $\left(-1 + 2i\right)\left(1 - 2i\right) = -1 + 2i + 2i - 4i^{2} = 3 + 4i$.

The answer need not come out in whole numbers. $\frac{{4 + i}}{{2 - 3i}} = \frac{\left(4 + i\right)\left(2 + 3i\right)}{{\left(2 - 3i\right)\left(2 + 3i\right)}} = \frac{{8 + 12i + 2i + 3i^{2}}}{{4 + 9}} = \frac{{5 + 14i}}{13}$, which is $\frac{5}{13} + \left(\frac{14}{13}\right)i$.

An imaginary denominator

For $\frac{{6 + 2i}}{{2i}}$, the conjugate of 2i is −2i. The bottom is $\left(2i\right)\left(-2i\right) = -4i^{2} = 4$, and the top is $\left(6 + 2i\right)\left(-2i\right) = -12i - 4i^{2} = 4 - 12i$. So the answer is $\frac{{4 - 12i}}{4} = 1 - 3i$. Check: $2i\left(1 - 3i\right) = 2i - 6i^{2} = 6 + 2i$.

The same move as rationalizing

A surd in a denominator is cleared the same way. $\frac{1}{{3 + \sqrt{2}}} = \frac{{3 - \sqrt{2}}}{{\left(3 + \sqrt{2}\right)\left(3 - \sqrt{2}\right)}} = \frac{{3 - \sqrt{2}}}{{9 - 2}} = \frac{{3 - \sqrt{2}}}{7}$. There the product of the pair loses the $\sqrt{2}$ because $\left(\sqrt{2}\right)^{2} = 2$; here it loses the i because $i^{2} = -1$.

The usual mistakes

Multiplying only the bottom. $\frac{{5 + 5i}}{10}$ is not equal to $\left(5 + 5i\right) / \left(3 + i\right)$: the top must be multiplied by 3 − i as well, which makes it 20 + 10i.

Losing the sign of $i^{2}$. $\left(3 + i\right)\left(3 - i\right) = 9 - i^{2} = 9 + 1 = 10$, not 9 − 1 = 8.

Using the conjugate of the top instead of the bottom. Multiplying $\frac{{5 + 5i}}{{3 + i}}$ by $\frac{{5 - 5i}}{{5 - 5i}}$ makes the bottom (3 + i)(5 − 5i) = 20 − 10i, which still has an i in it.

Changing the wrong sign. The conjugate of 3 + i is 3 − i, not −3 + i: only the sign of the imaginary part changes.

Dividing the parts separately. $\frac{{16 + 12i}}{{6 + 2i}}$ is not $\frac{16}{6} + \left(\frac{12}{2}\right)i$. That works only when the denominator is real.

Two branches in parallel

In the application below, two impedances in an AC circuit, 2 + 4i and 4 − 2i ohms, are joined in parallel, and their combined impedance is their product divided by their sum. The division is $\frac{{16 + 12i}}{{6 + 2i}}$, done by multiplying the top and the bottom by 6 − 2i.