Question

If $\frac{3}{2+\cos \theta +i\sin \theta }=a+ib$, then $(a-2{)}^2+b^2$ equals

• A. $-2$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $1$

The value of $\frac{3}{2+\cos \theta +i\sin \theta }=\frac{3}{2+\cos \theta +i\sin \theta }\times \frac{(2+\cos \theta )-i\sin \theta }{(2+\cos \theta )-i\sin \theta }$

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

$\Rightarrow a=\frac{6+3\sin \theta }{5+4\cos \theta },b=\frac{-3\sin \theta }{5+4\cos \theta }$

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

$=\frac{(-4-5\cos \theta {)}^2+9{\sin }^2\theta }{(5+4\cos \theta {)}^2}=\frac{(4+5\cos \theta {)}^2+9{\sin }^2\theta }{(5+4\cos \theta {)}^2}$

$=\frac{16+40\cos \theta +25{\cos }^2\theta +9{\sin }^2\theta }{(5+4\cos \theta {)}^2}$

$=\frac{16+40\cos \theta +16{\cos }^2\theta +9({\sin }^2\theta +{\cos }^2\theta )}{(5+4\cos \theta {)}^2}$

$=\frac{16+40\cos \theta +16{\cos }^2\theta +9}{(5+4\cos \theta {)}^2}$

$=\frac{16{\cos }^2\theta +40\cos \theta +25}{(5+4\cos \theta {)}^2}$

$=\frac{(4\cos \theta +5{)}^2}{(5+4\cos \theta {)}^2}=1$