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.
 A: Using polar form (with exponential notation for convenience),
$\frac{5i}{2+i}=\frac{5e^{i\frac{\pi}2}}{\sqrt 5 e^{i \arctan \frac 12}} = \sqrt 5 e^{i(\frac {\pi}2 - \arctan \frac 12)} = \sqrt 5 \cos (\frac {\pi}2 - \arctan \frac 12) + i \sqrt 5\sin (\frac {\pi}2 - \arctan \frac 12)$
Sketch the $1,2,\sqrt 5$ right triangle and determine that the cosine of the relevant angle (which is complementary to the angle with tangent $\frac 12$) is $\frac 1{\sqrt 5}$ and its sine is $\frac 2{\sqrt 5}$. The result above immediately simplifies to $1+2i$.
A: 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 $0\lt {\pi\over 2}-x\lt {\pi\over 2}$
A: You have\begin{align}\frac{5i}{2+i}&=\sqrt5\exp\left(\left(\frac\pi2-\arctan\left(\frac12\right)\right)i\right)\\&=\sqrt5\left(\cos\left(\frac\pi2-\arctan\left(\frac12\right)\right)+\sin\left(\frac\pi2-\arctan\left(\frac12\right)\right)i\right)\\&=\sqrt5\left(\sin\left(\arctan\left(\frac12\right)\right)+\cos\left(\arctan\left(\frac12\right)\right)i\right).\end{align}Now, note that\begin{align}\frac{\cos\left(\arctan\left(\frac12\right)\right)}{\sin\left(\arctan\left(\frac12\right)\right)}&=\cot\left(\arctan\left(\frac12\right)\right)\\&=\frac1{\tan\left(\arctan\left(\frac12\right)\right)}\\&=2.\end{align}Therefore,\begin{align}\frac{5i}{2+i}&=\sqrt5\exp\left(\arctan\left(2\right)i\right)\\&=\sqrt5\left(\frac1{\sqrt5}+\frac2{\sqrt5}i\right)\\&=1+2i.\end{align}
A: Even in complex analysis, if $z_1 \neq 0$, then 
$\displaystyle \frac{z_3}{z_1} = z_2 \iff (z_1 \times z_2) = z_3.$
$\displaystyle (2 + i) \times (1 + 2i) = 5i.$
A: We already have $ LHS =\sqrt 5 e^{i(\frac{\pi}{2}-x_1)}, x_1 =\arctan \frac{1}{2} \tag 1$.
Continue with $RHS=1+2i=\sqrt 5 e^{ix_2}, x_2=\arctan 2 \tag 2$.
$x_1, x_2$ are complementary angles, $x_2= \frac{\pi}{2}-x_1 \to LHS=RHS$
Note: polar form represented by exponential form for simplicity in (1)(2). There is no difference between two forms since $r$ and $\theta$ are same for LHS and RHS.
