How do you divide (4-9i)/(-6i)?

Jul 25, 2016

$\frac{3}{2} + \frac{2}{3} i$

Explanation:

We require to multiply the numerator and denominator of the fraction by the $\textcolor{b l u e}{\text{complex conjugate}}$ of the denominator. This will eliminate the i from the denominator, leaving a real number.

If z = a ± bi then the $\textcolor{b l u e}{\text{complex conjugate}}$ is

color(red)(|bar(ul(color(white)(a/a)color(black)(barz=a∓bi)color(white)(a/a)|)))
Note that the real part remains unchanged while the 'sign' of the imaginary part is reversed.

The complex number on the denominator is 0 - 6i , hence the conjugate is 0 + 6i = 6i

Multiply numerator and denominator by 6i

$\frac{4 - 9 i}{- 6 i} \times \frac{6 i}{6 i} = \frac{6 i \left(4 - 9 i\right)}{- 36 {i}^{2}} = \frac{24 i - 54 {i}^{2}}{- 36 {i}^{2}}$

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

$\Rightarrow \frac{4 - 9 i}{- 6 i} = \frac{24 i + 54}{36} = \frac{54}{36} + \frac{24}{36} i = \frac{3}{2} + \frac{2}{3} i$