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

Jul 28, 2017

The answer is $= \frac{5}{\sqrt{17}} \left(0.92 - 0.39 i\right)$

#### Explanation:

We start by writing numerator and denominator in polar form

$z = {z}_{1} / {z}_{2}$

The polar form of a complex number is

$z = r \left(\cos \theta + i \sin \theta\right) \ldots \ldots \ldots \ldots \ldots \ldots . .$(1)#

The numerator is

${z}_{1} = 3 + 4 i$

${r}_{1} = | {z}_{1} | = \sqrt{{\left(3\right)}^{2} + {\left(4\right)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

Therefore,

${z}_{1} = 5 \left(\frac{3}{5} + \frac{4}{5} i\right)$

Comparing this equation to equation $\left(1\right)$

$\cos \theta = \frac{3}{5}$ and $\sin \theta = \frac{4}{5}$

So,

we are in the quadrant $I$

$\theta = {53.13}^{\circ}$

The polar form is

${z}_{1} = 5 \left(\cos \left({53.13}^{\circ}\right) + i \sin \left({53.13}^{\circ}\right)\right) = 5 {e}^{53.13 i}$

The denominator is

${z}_{2} = 1 + 4 i$

${r}_{2} = | {z}_{2} | = \sqrt{{\left(1\right)}^{2} + {\left(4\right)}^{2}} = \sqrt{1 + 16} = \sqrt{17}$

Therefore,

${z}_{2} = \sqrt{17} \left(\frac{1}{\sqrt{17}} + \frac{4}{\sqrt{17}} i\right)$

Comparing this equation to equation $\left(1\right)$

$\cos \theta = \frac{1}{\sqrt{17}}$ and $\sin \theta = \frac{4}{\sqrt{17}}$

So,

we are in the quadrant $I$

$\theta = {75.96}^{\circ}$

The polar form is

${z}_{2} = \sqrt{17} \left(\cos \left({75.96}^{\circ}\right) + i \sin \left({75.96}^{\circ}\right)\right) = \sqrt{17} {e}^{75.96 i}$

Terefore,

$z = {z}_{1} / {z}_{2} = \frac{5 {e}^{53.13 i}}{\sqrt{17} {e}^{75.96 i}}$

$= \left(\frac{5}{\sqrt{17}}\right) {e}^{\left(53.13 - 75.96\right) i}$

$= \left(\frac{5}{\sqrt{17}}\right) {e}^{\left(- {22.83}^{\circ}\right) i}$

$= \frac{5}{\sqrt{17}} \left(\cos \left(- {22.83}^{\circ}\right) + i \sin \left(- {22.83}^{\circ}\right)\right)$

$= \frac{5}{\sqrt{17}} \left(0.92 - 0.39 i\right)$