Dividing Complex Numbers

An i in the denominator

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

Dividing by a complex number is not. $\frac{1}{2+i}$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 $\frac{2-i}{2-i}=1$(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)}$1/(2 + i) = (2 − i)/((2 + i)(2 − i)).

The denominator is a number times its conjugate, which is always real: $\left(2+i\right)\left(2-i\right)=4-i^2=4+1=5$(2 + i)(2 − i) = 4 − i² = 4 + 1 = 5. So $\frac{1}{2+i}=\frac{2-i}{5}=0.4-0.2i$1/(2 + i) = (2 − i)/5 = 0.4 − 0.2i.

The denominator (2 + i)(2 − i). The gold pieces 2i and −2i cancel, and $-i^2=+1$−i² = +1, so the denominator is the real number 5.

Check by multiplying back

If $\frac{1}{2+i}=0.4-0.2i$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$0.8 − 0.4i + 0.4i − 0.2i² = 0.8 + 0.2 = 1.

A full division

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

The bottom: $\left(3+i\right)\left(3-i\right)=9-i^2=9+1=10$(3 + i)(3 − i) = 9 − i² = 9 + 1 = 10. The top: $\left(5+5i\right)\left(3-i\right)=15-5i+15i-5i^2=15+10i+5=20+10i$(5 + 5i)(3 − i) = 15 − 5i + 15i − 5i² = 15 + 10i + 5 = 20 + 10i.

So $\left(5+5i\right)/\left(3+i\right)=\frac{20+10i}{10}=2+i$(5 + 5i) / (3 + i) = (20 + 10i)/10 = 2 + i. Now the denominator is real, each part is divided by it. Check: $\left(2+i\right)\left(3+i\right)=6+2i+3i+i^2=5+5i$(2 + i)(3 + i) = 6 + 2i + 3i + i² = 5 + 5i.

The new numerator (5 + 5i)(3 − i). Here the i-terms do not cancel: −5i + 15i = 10i. The gold piece is $-5i^2=+5$−5i² = +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 $\left(c+\text{di}\right)\left(c-\text{di}\right)=c^2+d^2$(c + di)(c − di) = c² + d², 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 $\left(3+4i\right)\left(1+2i\right)=3+6i+4i+8i^2=-5+10i$(3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i² = −5 + 10i and the bottom is 1 + 4 = 5, so the answer is −1 + 2i. Check: $\left(-1+2i\right)\left(1-2i\right)=-1+2i+2i-4i^2=3+4i$(−1 + 2i)(1 − 2i) = −1 + 2i + 2i − 4i² = 3 + 4i.

The answer need not come out in whole numbers. $\frac{4+i}{2-3i}=\frac{\left(4+i\right)\left(2+3i\right)}{\left(2-3i\right)\left(2+3i\right)}=\frac{8+12i+2i+3i^2}{4+9}=\frac{5+14i}{13}$(4 + i)/(2 − 3i) = (4 + i)(2 + 3i)/((2 − 3i)(2 + 3i)) = (8 + 12i + 2i + 3i²)/(4 + 9) = (5 + 14i)/13, which is $\frac{5}{13}+\left(\frac{14}{13}\right)i$5/13 + (14/13)i.

An imaginary denominator

For $\frac{6+2i}{2i}$(6 + 2i)/(2i), the conjugate of 2i is −2i. The bottom is $\left(2i\right)\left(-2i\right)=-4i^2=4$(2i)(−2i) = −4i² = 4, and the top is $\left(6+2i\right)\left(-2i\right)=-12i-4i^2=4-12i$(6 + 2i)(−2i) = −12i − 4i² = 4 − 12i. So the answer is $\frac{4-12i}{4}=1-3i$(4 − 12i)/4 = 1 − 3i. Check: $2i\left(1-3i\right)=2i-6i^2=6+2i$2i(1 − 3i) = 2i − 6i² = 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}$1/(3 + √2) = (3 − √2)/((3 + √2)(3 − √2)) = (3 − √2)/(9 − 2) = (3 − √2)/7. There the product of the pair loses the $\sqrt{2}$√2 because ${\left(\sqrt{2}\right)}^2=2$(√2)² = 2; here it loses the i because $i^2=-1$i² = −1.

The usual mistakes

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

Losing the sign of $i^2$i². $\left(3+i\right)\left(3-i\right)=9-i^2=9+1=10$(3 + i)(3 − i) = 9 − i² = 9 + 1 = 10, not 9 − 1 = 8.

Using the conjugate of the top instead of the bottom. Multiplying $\frac{5+5i}{3+i}$(5 + 5i)/(3 + i) by $\frac{5-5i}{5-5i}$(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. $\frac{16+12i}{6+2i}$(16 + 12i)/(6 + 2i) is not $\frac{16}{6}+\left(\frac{12}{2}\right)i$16/6 + (12/2)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 $\frac{16+12i}{6+2i}$(16 + 12i)/(6 + 2i), done by multiplying the top and the bottom by 6 − 2i.