# How do you divide (4+i)/(8i)?

Sep 11, 2016

$\frac{1}{8} - \frac{1}{2} i$

#### Explanation:

To divide this fraction we require the denominator to be a $\textcolor{b l u e}{\text{real number }}$and not a $\textcolor{b l u e}{\text{complex number}}$ as it is.

To achieve this multiply the numerator and denominator . in this case by i.

Since: $8 i \times i = 8 {i}^{2} = - 8 \leftarrow \text{ a real number}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Since this is a fraction, we must also multiply the numerator by i.

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

divide each term on the numerator by - 8

$\Rightarrow \frac{4 i}{- 8} - \frac{1}{- 8} = \frac{1}{8} - \frac{1}{2} i \text{ in standard form}$