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

Mar 1, 2018

$\frac{3 + 4 i}{1 + 4 i} = \frac{5}{\sqrt{17}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{8}{19}\right)$

#### Explanation:

Let us write the two complex numbers in polar coordinates and let them be

${z}_{1} = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and ${z}_{2} = {r}_{2} \left(\cos \beta + i \sin \beta\right)$

Here, if two complex numbers are ${a}_{1} + i {b}_{1}$ and ${a}_{2} + i {b}_{2}$ ${r}_{1} = \sqrt{{a}_{1}^{2} + {b}_{1}^{2}}$, ${r}_{2} = \sqrt{{a}_{2}^{2} + {b}_{2}^{2}}$ and $\alpha = {\tan}^{- 1} \left({b}_{1} / {a}_{1}\right)$, $\beta = {\tan}^{- 1} \left({b}_{2} / {a}_{2}\right)$

$\left\{{r}_{1} / {r}_{2}\right\} \left\{\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta}\right\}$ or

$\left\{{r}_{1} / {r}_{2}\right\} \left\{\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} \times \frac{\cos \beta - i \sin \beta}{\cos \beta - i \sin \beta}\right\}$

$\left({r}_{1} / {r}_{2}\right) \frac{\left(\cos \alpha \cos \beta + \sin \alpha \sin \beta\right) + i \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right)}{\left({\cos}^{2} \beta + {\sin}^{2} \beta\right)}$ or

$\left({r}_{1} / {r}_{2}\right) \cdot \left(\cos \left(\alpha - \beta\right) + i \sin \left(\alpha - \beta\right)\right)$ or

${z}_{1} / {z}_{2}$ is given by $\left({r}_{1} / {r}_{2} , \left(\alpha - \beta\right)\right)$

So for division complex number ${z}_{1}$ by ${z}_{2}$ , take new angle as $\left(\alpha - \beta\right)$ and modulus the ratio ${r}_{1} / {r}_{2}$ of the modulus of two numbers.

Here $3 + 4 i$ can be written as ${r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ where ${r}_{1} = \sqrt{{3}^{2} + {4}^{2}} = \sqrt{25} = 5$ and $\alpha = {\tan}^{- 1} \left(\frac{4}{3}\right)$

and $1 + 4 i$ can be written as ${r}_{2} \left(\cos \beta + i \sin \beta\right)$ where ${r}_{2} = \sqrt{{1}^{2} + {4}^{2}} = \sqrt{17}$ and $\beta = {\tan}^{- 1} 4$

and ${z}_{1} / {z}_{2} = \frac{5}{\sqrt{17}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = \alpha - \beta$

Hence, $\tan \theta = \tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{\frac{4}{3} - 4}{1 + \frac{4}{3} \times 4} = \frac{- \frac{8}{3}}{\frac{19}{3}} = - \frac{8}{19}$.

Hence, $\frac{3 + 4 i}{1 + 4 i} = \frac{5}{\sqrt{17}} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{8}{19}\right)$