Question

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

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

Answer 1

The correct option is B $4x-3$

$x+iy=\frac{3}{2+cos\theta +isin\theta }=\frac{3(2+cos\theta -isin\theta )}{(2+cos\theta {)}^2-i^{2}si{n}^2\theta }$

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

$=\frac{6+3cos\theta -3isin\theta }{4+co{s}^2\theta +4cos\theta +si{n}^2\theta }$

$=\frac{6+3cos\theta -3isin\theta }{5+4cos\theta }$

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

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

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

Hence, option B is correct.