# How do you divide ( i-3) / (2i+1) in trigonometric form?

Dec 17, 2017

See the explanation.

#### Explanation:

In trigonometric form,
$i - 3 = \sqrt{10} \left(\cos \alpha + i \sin \alpha\right)$, where
$\cos \alpha = - \frac{3}{\sqrt{10}} , \sin \alpha = \frac{1}{\sqrt{10}}$

$2 i + 1 = \sqrt{5} \left(\cos \beta + i \sin \beta\right)$
($\cos \beta = \frac{1}{\sqrt{5}} , \sin \beta = \frac{2}{\sqrt{5}}$).

So,
(i-3)/(2i+1)=(sqrt(10)(cosalpha+isinalpha))/(sqrt(5)(cosbeta+isinbeta)
$= \sqrt{2} \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$.

Then, use the trigonometric addition formulas.
cos(α-β)=cosαcosbeta+sinalphasinbeta
$= - \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{10}} \cdot \frac{2}{\sqrt{5}}$
$= - \frac{\sqrt{2}}{10}$

$\sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$
$= \frac{1}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} - \left(- \frac{3}{\sqrt{10}}\right) \cdot \frac{2}{\sqrt{5}}$
$= \frac{7 \sqrt{2}}{10}$

Therefore,
$\frac{i - 3}{2 i + 1} = \sqrt{2} \left\{- \frac{\sqrt{2}}{10} + i \cdot \left(\frac{7 \sqrt{2}}{10}\right)\right\}$
$= - \frac{1}{5} + \frac{7}{5} i$.