# How do you divide ( 3i-1) / (8i -4 ) in trigonometric form?

Mar 2, 2016

1/(2sqrt2)xxe^(i(alpha-beta), where $\alpha = {\tan}^{-} 1 \left(- 3\right)$ and $\beta = {\tan}^{-} 1 \left(- 2\right)$

#### Explanation:

$\left(a + b i\right)$ an be written in as sqrt(a^2+b^2)e^(i(tan^-1(b/a))

Hence, $\left(3 i - 1\right) = \left(- 1 + 3 i\right) = \sqrt{{1}^{2} + {3}^{2}} \left({e}^{i {\tan}^{-} 1 \left(- \frac{3}{1}\right)}\right) = \sqrt{10} {e}^{i {\tan}^{-} 1 \left(- 3\right)}$

Similarly, $\left(8 i - 4\right) = \left(- 4 + 8 i\right) = \sqrt{{4}^{2} + {8}^{2}} \left({e}^{i {\tan}^{-} 1 \left(- \frac{8}{4}\right)}\right) = \sqrt{80} {e}^{i {\tan}^{-} 1 \left(- 2\right)}$

Hence $\frac{3 i - 1}{8 i - 4} = \frac{\sqrt{10} {e}^{i {\tan}^{-} 1 \left(- 3\right)}}{\sqrt{80} {e}^{i {\tan}^{-} 1 \left(- 2\right)}}$

or sqrt(1/8)xxe^(i(tan^-1(-3))-(tan^-1(-2)) or

1/(2sqrt2)xxe^(i(alpha-beta), where $\alpha = {\tan}^{-} 1 \left(- 3\right)$ and $\beta = {\tan}^{-} 1 \left(- 2\right)$