Question

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

• A. 3x - 4
• B. 4x - 3
• C. 4x + 3
• D. None of these

Answer 1

Answer: (B)

$x+iy=\frac{3}{2+\cos \theta +i\sin \theta }$
$=\frac{3\left(2+\cos \theta -i\sin \theta \right)}{{\left(2+\cos \theta \right)}^2+{\sin }^2\theta }$
$=\frac{6+3\cos \theta -3i\sin \theta }{4+{\cos }^2\theta +4\cos \theta +{\sin }^2\theta }$
$=\frac{6+3\cos \theta -3i\sin \theta }{5+4\cos \theta }$
$=\left(\frac{6+3\cos \theta }{5+4\cos \theta }\right)+i\left(\frac{-3\sin \theta }{5+4\cos \theta }\right)$
On equating real and imaginary parts, we get
$x=\frac{3\left(2+\cos \theta \right)}{5+4\cos \theta }$
and $y=\frac{-3\sin \theta }{5+4\cos \theta }$
$\therefore x^2+y^2=\frac{9\left[4+{\cos }^2\theta +4\cos \theta +{\sin }^2\theta \right]}{{\left(5+4\cos \theta \right)}^2}$
$\left[4+{\cos }^2\theta +4\cos \theta +{\sin }^2\theta \right]$
$=\frac{9}{5+4\cos \theta }=4\left(\frac{6+3\cos \theta }{5+4{\cos }^2\theta }\right)-3$
$=4x-3$