# How do you divide  (5-2i)+(3-2i)?

Jan 5, 2017

$\frac{19}{13} + \frac{4}{13} i$

#### Explanation:

You can multiply both the terms by $3 + 2 i$ and get:

$\left(5 - 2 i\right) \textcolor{red}{\left(3 + 2 i\right)} \div \left(3 - 2 i\right) \textcolor{red}{\left(3 + 2 i\right)}$

Then, since

$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

expand:

$\left(15 + 10 i - 6 i - 4 {i}^{2}\right) \div \left(9 - 4 {i}^{2}\right)$

Then, since ${i}^{2} = - 1$, the expression becomes:

$\left(15 + 4 i + 4\right) \div \left(9 + 4\right)$

$\left(19 + 4 i\right) \div 13$

that's, in standard form:

$\frac{19}{13} + \frac{4}{13} i$