Show That $\frac{5i}{2+i}=1+2i$ With Exponents By writting the individual Factors on left in exponential form, perfrom needed operations and finally change back to rectangular coordinates 
show that $$ \frac{5i}{2+i}=1+2i$$

Attempt with exponents
$5i=5e^{i \pi /2}$. Also, $|2+i|=\sqrt{2^2+1^2}=\sqrt{5}$
let $\theta_1$ be angle from origin to point $(2,1)$ so $\tan(\theta _1 )=\frac{1}{2} \iff \theta_1 = \tan^{-1} \frac{1}{2}$ 
and we have
$$ \frac{5i}{2+i}=\frac{5e^{i \pi/2}}{\sqrt{5} e^{i \theta _1}}=\frac{5}{\sqrt{5} } e^{i (\pi/2 -\theta _1 )} $$
$\theta _1$ is giving me some trouble. Can reverse engineer what $e^{i (\pi/2 -\theta _1 )}$  is suppose to be. But wolfram gives me jumble up answer. I am wondering is there a trick. Not sure how to deal with it. I like to draw the unit cirlcle and somehow get what it is but i cant.

can show that it works in euclidian 
$$\frac{z_1}{z_2}= \frac{(x_1 x_2 +y_1 y_2,-x_1 y_2 +y_1 y_2  )}{x_2 ^2 +y_2 ^2 } $$
in our case $z_1=(0,5)$ so $x_1 =0 ,y_1 =5$. $z_2=(2,1)$ so $x_2 =2,y_2 =1$ 
and we get that  
$$\begin{aligned}
    \frac{5i}{2+i} &= \frac{(0*2+5*1,-*0*1+5*2)}{2^2 +1^2}
\\  &=\frac{(5,10)}{5}=(5/5,10/5)=(1,2)=1+2i \end{aligned}$$
 A: You can use:
$$e^{i\theta}=\cos \theta + i\sin \theta$$
Then $$\sqrt{5} e^{i(\frac {\pi}2 +\theta_1)}=\sqrt 5 \left(\cos \left( \frac {\pi}2 +\theta_1\right) +i\sin \left( \frac {\pi}2 +\theta_1\right) \right)$$
Then use trigonometric identities and find the value explicitly.
A: Use, when $\text{q}\space\wedge\space\text{z}\in\mathbb{C}$:


*

*$$\frac{\text{q}}{\text{z}}=\frac{\text{q}\overline{\text{z}}}{\text{z}\overline{\text{z}}}=\frac{\text{q}\overline{\text{z}}}{|\text{z}|^2}=\frac{\Re[\text{z}]\Re[\text{q}]+\Im[\text{z}]\Im[\text{q}]+\left(\Re[\text{z}]\Im[\text{q}]-\Re[\text{z}]\Im[\text{q}]\right)i}{\Re^2[\text{z}]+\Im^2[\text{z}]}$$

*With exponentials (where $k\in\mathbb{Z}$):
$$\frac{\text{q}}{\text{z}}=\left|\frac{\text{q}}{\text{z}}\right|e^{\left(\arg\left(\frac{\text{q}}{\text{z}}\right)+2\pi k\right)i}=\frac{\left|\text{q}\right|}{\left|\text{z}\right|}e^{\left(\arg\left(\text{q}\right)-\arg\left(\text{z}\right)+2\pi k\right)i}$$


Now:


*

*When $\text{q}\in\mathbb{C}$ with $\Re[\text{q}]=0$ and $\Im[\text{q}]>0$:
$$\arg(\text{q})=\frac{\pi}{2}$$

*When $\text{z}\in\mathbb{C}$ with $\Re[\text{z}]>0$ and $\Im[\text{z}]>0$:
$$\arg(\text{z})=\arctan\left(\frac{\Im[\text{z}]}{\Re[\text{z}]}\right)$$

*For $|\text{q}|$, we get:
$$|\text{q}|=\sqrt{\Re^2[\text{q}]+\Im^2[\text{q}]}$$

*For $|\text{z}|$, we get:
$$|\text{z}|=\sqrt{\Re^2[\text{z}]+\Im^2[\text{z}]}$$

*So, for $\frac{\text{q}}{\text{z}}$ we get (with the assumptions from above and using exponentials):
$$\frac{\text{q}}{\text{z}}=\frac{\sqrt{\Re^2[\text{q}]+\Im^2[\text{q}]}}{\sqrt{\Re^2[\text{z}]+\Im^2[\text{z}]}}\cdot e^{\left(\frac{\pi}{2}-\arctan\left(\frac{\Im[\text{z}]}{\Re[\text{z}]}\right)+2\pi k\right)i}$$


So, we get:


*

*$$\Re\left[\frac{\text{q}}{\text{z}}\right]=\frac{\sqrt{\Re^2[\text{q}]+\Im^2[\text{q}]}}{\sqrt{\Re^2[\text{z}]+\Im^2[\text{z}]}}\cdot\cos\left(\frac{\pi}{2}-\arctan\left(\frac{\Im[\text{z}]}{\Re[\text{z}]}\right)+2\pi
   k\right)$$

*$$\Im\left[\frac{\text{q}}{\text{z}}\right]=\frac{\sqrt{\Re^2[\text{q}]+\Im^2[\text{q}]}}{\sqrt{\Re^2[\text{z}]+\Im^2[\text{z}]}}\cdot\sin\left(\frac{\pi}{2}-\arctan\left(\frac{\Im[\text{z}]}{\Re[\text{z}]}\right)+2\pi
   k\right)$$

