# How do you divide (-i+2) / (2i+4) in trigonometric form?

Apr 28, 2016

$\frac{- i + 2}{2 i + 4} = \frac{1}{2} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{4}{3}\right)$

#### 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 $\alpha = {\tan}^{- 1} \left({b}_{1} / {a}_{1}\right)$, $\beta = {\tan}^{- 1} \left({b}_{2} / {a}_{2}\right)$

Their division leads us to

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

and $2 i + 4$ can be written as ${r}_{2} \left(\cos \beta + i \sin \beta\right)$ where ${r}_{2} = \sqrt{{4}^{2} + {2}^{2}} = \sqrt{20} = 2 \sqrt{5}$ and $\beta = {\tan}^{- 1} \left(\frac{2}{4}\right) = {\tan}^{- 1} \left(\frac{1}{2}\right)$

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

Hence, $\tan \theta = \tan \left(\alpha - \beta\right) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{\left(- \frac{1}{2}\right) - \left(\frac{1}{2}\right)}{1 + \left(- \frac{1}{2}\right) \times \left(\frac{1}{2}\right)} = \frac{- 1}{\frac{3}{4}} = - \frac{4}{3}$.

Hence, $\frac{- i + 2}{2 i + 4} = \frac{1}{2} \left(\cos \theta + i \sin \theta\right)$, where $\theta = {\tan}^{- 1} \left(- \frac{4}{3}\right)$

Jun 30, 2016

$\frac{- i + 2}{2 i + 4} = \frac{1}{2} \left(\cos \theta + \sin \theta\right)$ ,where $\theta = {\tan}^{-} 1 \left(- \frac{4}{3}\right)$

#### Explanation:

$\frac{- i + 2}{2 i + 4}$

$= \frac{1}{2} \cdot \frac{\left(2 - i\right) \left(2 - i\right)}{\left(2 + i\right) \left(2 - i\right)}$

$= \frac{1}{2} \cdot \frac{{2}^{2} + {i}^{2} - 2 \cdot 2 i}{{2}^{2} - {i}^{2}}$

$= \frac{1}{2} \cdot \frac{4 - 1 - 4 i}{4 - \left(- 1\right)}$

$= \frac{1}{10} \left(3 - 4 i\right)$

Now $\sqrt{{3}^{2} + {4}^{2}} = 5$,So

The given expression

$= \frac{1}{2} \left(\frac{3}{5} - \frac{4}{5} i\right)$

Now if we take $\frac{3}{5} = \cos \theta$ and $- \frac{4}{5} = \sin \theta$
i.e.$\tan \theta = - \frac{4}{3}$
then we can write

$\frac{- i + 2}{2 i + 4} = \frac{1}{2} \left(\cos \theta + \sin \theta\right)$ ,where $\theta = {\tan}^{-} 1 \left(- \frac{4}{3}\right)$