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  4. If x+iy = (3/2+ cos θ +i sinθ), then x2 + y2 is equal to

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

VITEEEVITEEE 2014

Solution:

$x + iy = \frac{3}{2+\cos \theta+i \sin\theta}$
$= \frac{3\left(2 +\cos \theta- i \sin\theta\right)}{\left(2+ \cos\theta\right)^{2} + \sin^{2} \theta}$
$= \frac{6+3\cos \theta-3i \sin\theta}{4+\cos^{2} \theta+4 \cos \theta + \sin^{2} \theta}$
$ = \frac{6+3 \cos\theta-3 i \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} $
$\therefore x^{2} +y^{2} = \frac{9\left[4+ \cos^{2 } \theta+4 \cos\theta+\sin^{2} \theta\right]}{\left(5+4 \cos\theta\right)^{2}}$
$\left[4+\cos ^{2} \theta+4 \cos \theta+\sin ^{2} \theta\right]$
$= \frac{9}{5+4 \cos\theta} = 4 \left(\frac{6+3 \cos\theta}{5+4 \cos^{2} \theta}\right) - 3$
$ = 4 x - 3 $