Prove $\frac{5i}{2+i}=1+2i$ Since $5i=5e^{i \frac{\pi}{2}}$
And $2+i=\sqrt{5}e^{ix_1}$ where $x_1=\arctan \frac{1}{2}$, we have $\frac{5i}{2+i}=\sqrt{5}[\cos(\frac{\pi}{2}-x_1)+i\sin(\frac{\pi}{2}-x_1)]$
Now from here how do I continue?
P.s:I know how to solve this if it wasn't asked to solve using polar forms.

Hint
$$\begin{align}
\cos\left({\pi \over 2}-x\right)&=\sin{x}\\
\sin\left({\pi \over 2}-x\right)&=\cos{x}\\
\tan\left({\pi \over 2}-x\right)&={1\over\tan{x}}
\end{align}$$
So now let's develop the hint from where you left it i.e
$${5i\over 1+2i}=\sqrt{5}\left(\cos\left({\pi\over 2}-x\right)+i\sin\left({\pi\over 2}-x\right)\right)=a+ib$$
So $a^2+b^2=5$ and
$${b\over a}=\tan\left({\pi\over 2}-x\right)={1\over \tan{x}}=2$$
And so $5a^2=5$ and $a=\pm 1$, $b=\pm 2$ The negative solutions are to be excluded  because $0\lt {\pi\over 2}-x\lt {\pi\over 2}$
