# How do you divide ( -i+1) / (2i +10 ) in trigonometric form?

Apr 6, 2018

$\frac{- i + 1}{2 i + 10} = \frac{1}{2} \left(\frac{2}{13} - i \frac{3}{13}\right)$

#### Explanation:

A complex number $z$ is of the form

$z = a + b i$

We define the Polar coordinates of $z$ to be $\left(r , \theta\right)$, as seen in the image below.

From this diagram we get some more properties:

$r = \sqrt{{a}^{2} + {b}^{2}}$

$\sin \theta = \frac{b}{r} \implies b = r \sin \theta$
$\cos \theta = \frac{a}{r} \implies a = r \cos \theta$

If we substitute $b$ and $a$ into the definition of a complex number, we have

$z = r \cos \theta + i r \sin \theta = r \left(\cos \theta + i \sin \theta\right)$

Our trigonometric sum resembles $\textcolor{red}{\text{Euler's identity}}$:

e^(icolor(red)alpha)=coscolor(red)alpha+isincolor(Red)(alpha

Thus,

$z = r {e}^{i \theta}$

In our case, it'd be easier to write them in this exponential form then transform it into trigonometric form.

Let color(blue)(z_1 = -i+1 and color(blue)(z_2 = 2i+10.

We do not need to find ${\theta}_{1}$ and ${\theta}_{2}$ now, so we will let them as that.

${r}_{1} = \sqrt{{a}_{1}^{2} + {b}_{1}^{2}} = \sqrt{{1}^{2} + {\left(- 1\right)}^{2}} = \sqrt{2}$
${r}_{2} = \sqrt{{a}_{2}^{2} + {b}_{2}^{2}} = \sqrt{100 + 4} = \sqrt{104} = 2 \sqrt{26}$

$\therefore {z}_{1} / {z}_{2} = \frac{{r}_{1} {e}^{i {\theta}_{1}}}{{r}_{2} {e}^{i {\theta}_{2}}}$

z_1/z_2 = sqrt2/(2sqrt26) * e^(i(theta_1-theta_2)

z_1/z_2 = 1/(2sqrt13) * e^(i(theta_1-theta_2)

We can still apply Euler's identity to ${e}^{i \left({\theta}_{1} - {\theta}_{2}\right)}$. We have:

${e}^{i \left({\theta}_{1} - {\theta}_{2}\right)} = \cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left({\theta}_{1} - {\theta}_{2}\right)$

The $\textcolor{b l u e}{\text{Difference formula}}$ for cosine and sine is, as follows:

$\cos \left(a - b\right) = \cos a \cos b + \sin a \sin b$
$\sin \left(a - b\right) = \sin a \cos b - \cos a \sin b$

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

From the properties we got earlier, we know:

$\cos {\theta}_{1} = \frac{1}{\sqrt{2}}$
$\sin {\theta}_{1} = - \frac{1}{\sqrt{2}}$

$\cos {\theta}_{2} = \frac{5}{\sqrt{26}}$
$\sin {\theta}_{2} = \frac{1}{\sqrt{26}}$

After we calculate the values we needed, we reach this:

$\cos \left({\theta}_{1} - {\theta}_{2}\right) = \frac{2}{\sqrt{13}}$
$\sin \left({\theta}_{1} - {\theta}_{2}\right) = - \frac{3}{\sqrt{13}}$

Finally, we get:

${z}_{1} / {z}_{2} = \frac{1}{2 \sqrt{13}} \left(\frac{2}{\sqrt{13}} - i \frac{3}{\sqrt{13}}\right)$

${z}_{1} / {z}_{2} = \frac{1}{2} \left(\frac{2}{13} - i \frac{3}{13}\right)$

$\therefore$

$\frac{- i + 1}{2 i + 10} = \frac{1}{2} \left(\frac{2}{13} - i \frac{3}{13}\right)$