# How do you write the following quotient in standard form (8-7i)/(1-2i)?

Aug 3, 2016

22/5+9/5i

#### Explanation:

Since $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2} \mathmr{and} {i}^{2} = - 1$,

you can multiply numerator and denominator of the fraction by the expression $\left(1 + 2 i\right)$ and have:

$\frac{\left(8 - 7 i\right) \left(1 + 2 i\right)}{\left(1 - 2 i\right) \left(1 + 2 i\right)} = \frac{8 + 16 i - 7 i - 14 {i}^{2}}{1 - 4 {i}^{2}}$

$\frac{8 + 9 i + 14}{1 + 4} = \frac{22 + 9 i}{5} = \frac{22}{5} + \frac{9}{5} i$

in standard form