# How do you write the following in trigonometric form and perform the operation given (3+4i)/(1-sqrt3i)?

Nov 13, 2016

$z = \frac{3}{4} - \sqrt[2]{3} + i \left(\frac{3}{4} \sqrt[2]{3} + 1\right) = \frac{5}{2} \left(\cos \theta + i \sin \theta\right)$
with $\theta = \arctan \left(\frac{\frac{25}{16} \sqrt[2]{3} + 3}{- \frac{39}{16}}\right)$

#### Explanation:

let's get rid of complex denominator...
$z = \frac{3 + 4 i}{1 - i \sqrt[2]{3}} \frac{1 + i \sqrt[2]{3}}{1 + i \sqrt[2]{3}} = \frac{3 + i 3 \sqrt[2]{3} + i 4 - 4 \sqrt[2]{3}}{4}$
$z = \frac{3}{4} - \sqrt[2]{3} + i \left(\frac{3}{4} \sqrt[2]{3} + 1\right)$ that in trigonometric form is $z = \left\mid z \right\mid \left(\cos \theta + i \sin \theta\right)$
$\left\mid z \right\mid = \sqrt[2]{{\left(\frac{3}{4} - \sqrt[2]{3}\right)}^{2} + {\left(\frac{3}{4} \sqrt[2]{3} + 1\right)}^{2}}$
$\left\mid z \right\mid = \sqrt[2]{\frac{9}{16} + 3 - \frac{3}{2} \sqrt[2]{3} + \frac{27}{16} + 1 + \frac{3}{2} \sqrt[2]{3}} =$
$\left\mid z \right\mid = \sqrt[2]{\frac{36 + 64}{16}} = \frac{10}{4} = \frac{5}{2}$
the angle is $\theta = \arctan \left(\frac{\frac{3}{4} \sqrt[2]{3} + 1}{\frac{3}{4} - \sqrt[2]{3}}\right)$
$\theta = \arctan \left(\frac{\frac{3}{4} \sqrt[2]{3} + 1}{\frac{3}{4} - \sqrt[2]{3}} \cdot \frac{\frac{3}{4} + \sqrt[2]{3}}{\frac{3}{4} + \sqrt[2]{3}}\right) =$

$\theta = \arctan \left(\frac{\frac{9}{16} \sqrt[2]{3} + \frac{9}{4} + \frac{3}{4} + \sqrt[2]{3}}{\frac{9}{16} - 3}\right)$
$\theta = \arctan \left(\frac{\frac{25}{16} \sqrt[2]{3} + 3}{- \frac{39}{16}}\right)$