How do you divide (2i)/(3-9i)?

Sep 20, 2016

$- \frac{1}{5} + \frac{1}{15} i$

Explanation:

Before we can divide, we require the denominator to be a real number.
This is achieved by multiplying the denominator by it's $\textcolor{b l u e}{\text{conjugate}}$

$\text{If "z=x±yi" is a complex number then it's conjugate is}$

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

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

Hence the conjugate of 3-9i is 3 + 9i

and if we multiply them.

$\left(3 - 9 i\right) \left(3 + 9 i\right) = 9 + 27 i - 27 i - 81 {i}^{2} = 9 - 81 {i}^{2}$

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

Then $9 - 81 {i}^{2} = 90 \leftarrow \text{ a real number}$

Since we have a fraction then the numerator/denominator must be multiplied by the complex conjugate.

$\frac{2 i}{3 - 9 i} = \frac{2 i \left(3 + 9 i\right)}{\left(3 - 9 i\right) \left(3 + 9 i\right)} = \frac{6 i + 18 {i}^{2}}{90} = \frac{- 18 + 6 i}{90}$

$= \frac{- 18}{90} + \frac{6}{90} i = - \frac{1}{5} + \frac{1}{15} i$