# How do you divide ( i+3) / (-3i +7 ) in trigonometric form?

Feb 22, 2018

$0.311 + 0.275 i$

#### Explanation:

First I will rewrite the expressions in the form of $a + b i$

$\frac{3 + i}{7 - 3 i}$

For a complex number $z = a + b i$, $z = r \left(\cos \theta + i \sin \theta\right)$, where:

• $r = \sqrt{{a}^{2} + {b}^{2}}$
• $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let's call $3 + i$ ${z}_{1}$ and $7 - 3 i$ ${z}_{2}$.

For ${z}_{1}$:

${z}_{1} = {r}_{1} \left(\cos {\theta}_{1} + i \sin {\theta}_{1}\right)$

${r}_{1} = \sqrt{{3}^{2} + {1}^{2}} = \sqrt{9 + 1} = \sqrt{10}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{1}{3}\right) = {0.32}^{c}$

${z}_{1} = \sqrt{10} \left(\cos \left(0.32\right) + i \sin \left(0.32\right)\right)$

For ${z}_{2}$:

${z}_{2} = {r}_{2} \left(\cos {\theta}_{2} + i \sin {\theta}_{2}\right)$

${r}_{2} = \sqrt{{7}^{2} + {\left(- 3\right)}^{2}} = \sqrt{58}$

${\theta}_{2} = {\tan}^{-} 1 \left(- \frac{3}{7}\right) = - {0.40}^{c}$

However, since $7 - 3 i$ is in quadrant 4, we need to get a positive angle equivalent (the negative angle goes clockwise around the circle, and we need an anticlockwise angle).

To get a positive angle equivalent, we add $2 \pi$, ${\tan}^{-} 1 \left(- \frac{3}{7}\right) + 2 \pi = {5.88}^{c}$

${z}_{2} = \sqrt{58} \left(\cos \left(5.88\right) + i \sin \left(5.88\right)\right)$

For ${z}_{1} / {z}_{2}$:

${z}_{1} / {z}_{2} = {r}_{1} / {r}_{2} \left(\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left({\theta}_{1} - {\theta}_{2}\right)\right)$

$\textcolor{w h i t e}{{z}_{1} / {z}_{2}} = \frac{\sqrt{10}}{\sqrt{58}} \left(\cos \left[{\tan}^{-} 1 \left(\frac{1}{3}\right) - \left({\tan}^{-} 1 \left(- \frac{3}{7}\right) + 2 \pi\right)\right] + i \sin \left[{\tan}^{-} 1 \left(\frac{1}{3}\right) - \left({\tan}^{-} 1 \left(- \frac{3}{7}\right) + 2 \pi\right)\right]\right)$

$\textcolor{w h i t e}{{z}_{1} / {z}_{2}} = \frac{\sqrt{145}}{29} \left(\cos \left[{\tan}^{-} 1 \left(\frac{1}{3}\right) - {\tan}^{-} 1 \left(- \frac{3}{7}\right) - 2 \pi\right] + i \sin \left[{\tan}^{-} 1 \left(\frac{1}{3}\right) - {\tan}^{-} 1 \left(- \frac{3}{7}\right) - 2 \pi\right]\right)$

$\textcolor{w h i t e}{{z}_{1} / {z}_{2}} = \frac{\sqrt{145}}{29} \left(\cos \left(- 5.56\right) + i \sin \left(- 5.56\right)\right)$

$\textcolor{w h i t e}{{z}_{1} / {z}_{2}} = \frac{\sqrt{145}}{29} \cos \left(- 5.56\right) + i \frac{\sqrt{145}}{29} \sin \left(- 5.56\right)$

$\textcolor{w h i t e}{{z}_{1} / {z}_{2}} = 0.311 + 0.275 i$

Proof:
$\frac{3 + i}{7 - 3 i} \cdot \frac{7 + 3 i}{7 + 3 i} = \frac{\left(3 + i\right) \left(7 + 3 i\right)}{\left(7 - 3 i\right) \left(7 + 3 i\right)} = \frac{21 + 7 i + 9 i + 3 {i}^{2}}{49 + 21 i - 21 i - 9 {i}^{2}} = \frac{21 + 16 i + 3 {i}^{2}}{49 - 9 {i}^{2}}$

${i}^{2} = - 1$

$= \frac{21 + 16 i - 3}{49 + 9} = \frac{18 + 16 i}{58} = \frac{9}{29} + \frac{8}{29} i \approx 0.310 + 0.275 i$