# How do you simplify  (1+3i) div (5-3i) ?

Nov 22, 2015

$- \frac{1}{9} + \frac{1}{2} i$

#### Explanation:

Write as $\frac{1 + 3 i}{5 - 3 i}$. Multiply by the complex conjugate of the denominator.

$\frac{1 + 3 i}{5 - 3 i} \left(\frac{5 + 3 i}{5 + 3 i}\right) = \frac{5 + 3 i + 15 i + 9 {i}^{2}}{25 \cancel{- 15 i + 15 i} - 9 {i}^{2}} = \frac{5 + 18 i + 9 {i}^{2}}{25 - 9 {i}^{2}}$

Recall that $i = \sqrt{- 1}$, so color(red)(i^2=-1.

$\frac{5 + 18 i + 9 \left(- 1\right)}{25 - 9 \left(- 1\right)} = \frac{- 4 + 18 i}{36} = - \frac{1}{9} + \frac{1}{2} i$

Note that the answer is written in the $a + b i$ form of a complex number.