# How do you simplify (-5-3i)/(4i) and write the complex number in standard form?

Jul 15, 2016

#### Answer:

$- \frac{3}{4} + \frac{5}{4} i$

#### Explanation:

To simplify, we require to multiply the numerator and denominator by the $\textcolor{b l u e}{\text{complex conjugate}}$ of the denominator.

Given a complex number z = x ± yi then the complex conjugate is.

color(red)(|bar(ul(color(white)(a/a)color(black)(barz=x∓yi)color(white)(a/a)|)))

Note that the real part remains unchanged while the $\textcolor{b l u e}{\text{sign}}$ of the imaginary part is reversed.

Example: z = 5 + 4i then complex conjugate is $\overline{z} = 5 - 4 i$

Here z = 4i = 0 + 4i in standard form.Thus the complex conjugate is - 4i

Multiply numerator and denominator by - 4i

$\Rightarrow \left(\left(- 5 - 3 i\right) \textcolor{red}{\text{-4i")/(4ixxcolor(red)"-4i}}\right) = \frac{20 i + 12 {i}^{2}}{- 16 {i}^{2}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

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

$\Rightarrow \frac{20 i + 12 {i}^{2}}{- 16 {i}^{2}} = \frac{20 i - 12}{16} = \frac{- 12}{16} + \frac{20}{16} i$

$\Rightarrow \frac{- 5 - 3 i}{4 i} = - \frac{3}{4} + \frac{5}{4} i \text{ in standard form}$