Discussion Forum

If $x+iy=\frac{3}{2+\cos \theta+i \sin \theta}$, then $x^{2}+y^{2}$ is equal to 

Answer:

Explanation:

$x+iy$

   $=\frac{3}{2+\cos \theta+i \sin \theta}=\frac{3(2+\cos \theta-i \sin \theta)}{(2+\cos \theta)^{2}+\sin^{2} \theta}$

  $=\frac{6+3\cos \theta-3 i \sin \theta}{4+\cos^{2} \theta+4 \cos \theta+\sin^{2} \theta}$

$\frac{6+3 \cos \theta-3i \sin \theta}{5+4 \cos \theta}$

   $=\left(\frac{6+3\cos \theta}{5+4 \cos \theta}\right)+\left(\frac{-3 \sin \theta}{5+4 \cos \theta}\right)$

 On equating  real and imaginary  parts, we get

$x= \frac{3(2+\cos \theta)}{5+4\cos \theta}$

and $y=\frac{-3 \sin \theta}{5+4 \cos \theta}$

$\therefore$   $x^{2}+y^{2}= \frac{9[4+\cos^{2} \theta+4 \cos \theta+\sin^{2}\theta]}{(5+4 \cos \theta)^{2}}$

$=\frac{9}{5+4 \cos \theta}=4\left(\frac{6+3 \cos \theta}{5+4 \cos^{2} \theta}\right)-3$

   =4x-3