# How do you write the following in trigonometric form and perform the operation given 5/(2+3i)?

Jun 3, 2017

$\frac{5}{2 + 3 i} = \frac{10}{13} - i \frac{15}{13}$

#### Explanation:

Here two numbers are involved

One is $5$ which is easy to write in trigonometric form as

$5 = 5 \left(1 + i 0\right) = 5 \left(\cos {0}^{\circ} + i \sin {0}^{\circ}\right)$ or $5 {e}^{i 0}$

For $2 + 3 i$, we know $\sqrt{{2}^{2} + {3}^{2}} = \sqrt{4 + 9} = \sqrt{13}$

Hence $2 + 3 i = \sqrt{13} \left(\frac{2}{\sqrt{13}} + i \frac{3}{\sqrt{13}}\right)$

Now let $\sin \alpha = \frac{3}{\sqrt{13}}$ and $\cos \alpha = \frac{2}{\sqrt{13}}$, then

$2 + 3 i = \sqrt{13} \left(\cos \alpha + i \sin \alpha\right) = 5 {e}^{i \alpha}$

and 5/(2+3i)=(5e^(i0))/(sqrt13e^(ialpha)

= $\frac{5}{\sqrt{13}} \times {e}^{0 - i \alpha} = \frac{5}{\sqrt{13}} {e}^{- i \alpha}$

= $\frac{5}{\sqrt{13}} \left(\cos \left(- \alpha\right) + i \sin \left(- \alpha\right)\right)$

= $\frac{5}{\sqrt{13}} \left(\cos \alpha - i \sin \alpha\right)$

= $\frac{5}{\sqrt{13}} \left(\frac{2}{\sqrt{13}} - i \times \frac{3}{\sqrt{13}}\right)$

= $\frac{5}{13} \left(2 - 3 i\right) = \frac{10}{13} - i \frac{15}{13}$