# How do you divide (i+2) / (9i+14) in trigonometric form?

Apr 19, 2018

$0.134 - 0.015 i$

#### Explanation:

For a complex number $z = a + b i$ it can be represented as $z = r \left(\cos \theta + i \sin \theta\right)$ where $r = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

$\frac{2 + i}{14 + 9 i} = \frac{\sqrt{{2}^{2} + {1}^{2}} \left(\cos \left({\tan}^{-} 1 \left(\frac{1}{2}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{1}{2}\right)\right)\right)}{\sqrt{{14}^{2} + {9}^{2}} \left(\cos \left({\tan}^{-} 1 \left(\frac{9}{14}\right)\right) + i \sin \left({\tan}^{-} 1 \left(\frac{9}{14}\right)\right)\right)} \approx \frac{\sqrt{5} \left(\cos \left(0.46\right) + i \sin \left(0.46\right)\right)}{\sqrt{277} \left(\cos \left(0.57\right) + i \sin \left(0.57\right)\right)}$

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

${z}_{1} / {z}_{2} = \frac{\sqrt{5}}{\sqrt{277}} \left(\cos \left(0.46 - 0.57\right) + i \sin \left(0.46 - 0.57\right)\right) = \frac{\sqrt{1385}}{277} \left(\cos \left(- 0.11\right) + i \sin \left(- 0.11\right)\right) \approx \frac{\sqrt{1385}}{277} \left(0.99 - 0.11 i\right) \approx 0.134 - 0.015 i$

Proof:
$\frac{2 + i}{14 + 9 i} \cdot \frac{14 - 9 i}{14 - 9 i} = \frac{28 - 4 i + 9}{{14}^{2} + {9}^{2}} = \frac{37 - 4 i}{277} \approx 0.134 - 0.014 i$