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If $x + iy = \frac{3}{{2 + \cos \theta + i\sin \theta }},$then ${x^2} + {y^2}$ is equal to
$3x - 4$
$4x - 3$
$4x + 3$
None of these
Solution
(b) If $x + iy = \frac{3}{{2 + \cos \theta + i\sin \theta }}$
$ = \frac{{3(2 + \cos \theta – i\sin \theta )}}{{{{(2 + \cos \theta )}^2} + {{\sin }^2}\theta }} = \frac{{6 + 3\cos \theta – 3i\sin \theta }}{{4 + {{\cos }^2}\theta + 4\cos \theta + {{\sin }^2}\theta }}$
$ = \left[ {\frac{{6 + 3\cos \theta }}{{5 + 4\cos \theta }}} \right] + i\,\left[ {\frac{{ – 3\sin \theta }}{{5 + 4\cos \theta }}} \right]$
==> $x = \frac{{3(2 + \cos \theta )}}{{5 + 4\cos \theta }},y = \frac{{ – 3\sin \theta }}{{5 + 4\cos \theta }}$
${x^2} + {y^2} = \frac{9}{{{{(5 + 4\cos \theta )}^2}}}$ $[4 + {\cos ^2}\theta + 4\cos \theta + {\sin ^2}\theta ]$
$ = \frac{9}{{5 + 4\cos \theta }} = 4\left[ {\frac{{6 + 3\cos \theta }}{{5 + 4\cos \theta }}} \right] – 3 = 4x – 3$
Trick : $x + iy = \frac{{3(2 + \cos \theta – i\sin \theta )}}{{(2 + \cos \theta + i\sin \theta )(2 + \cos \theta – i\sin \theta )}}$
Let $y = 0$, then $\sin \theta = 0$i.e., $\theta = 0$.
Now put $x = 1$ then ${x^2} + {y^2} = {1^2} + 0 = 1$.
Also option (b) gives $4(1) -3=1.$
