If ${(\frac{3}{2}+\frac{i\sqrt{3}}{2})}^{50}=3^{25}(x-iy),$ where $x,y\in R$ and $i=\sqrt{-1},$ find the ordered pair of $(x,y).$ 

$\frac{3}{2}+\frac{i\sqrt{3}}{2}=\sqrt{3}\left(\frac{\sqrt{3}+i}{2}\right)=\frac{\sqrt{3}}{i}\left(\frac{i\sqrt{3}+i^2}{2}\right)$ 
$=-i\sqrt{3}\left(\frac{-1+i\sqrt{3}}{2}\right)=-i\sqrt{3}\omega$ 
$\therefore {(\frac{3}{2}+\frac{i\sqrt{3}}{2})}^{50}={(-i\sqrt{3}\omega )}^{50}=i^{50}{.3}^{25}.{\omega }^{50}=-{1.3}^{25}.{\omega }^2=-3^{25}.\left(\frac{1-i\sqrt{3}}{2}\right)$ 
$=3^{25(\frac{1}{2}+\frac{i\sqrt{3}}{2})}=3^{25}(x-iy)$ [given]
$\therefore x=\frac{1}{2},y=-\frac{\sqrt{3}}{2}\Rightarrow$Ordered pair is $(\frac{1}{2},-\frac{\sqrt{3}}{2}).$