Question

If $\frac{3}{2+\cos \theta +i\sin \theta }=x+iy$ then $(x-1)(x-3)=$

• A. $y^2$
• B. $-y^2$
• C. $0$
• D. $1$

Answer 1

The correct option is A $y^2$

$x+iy=\frac{3}{2+\cos \theta +i\sin \theta }$

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

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

$x+iy=\frac{3(2+\cos \theta )}{5+4\cos \theta }+i.\frac{-3\sin \theta }{5+4\cos \theta }$

$\therefore x=\frac{3(2+\cos \theta )}{5+4\cos \theta };y=\frac{-3\sin \theta }{5+4\cos \theta }$

Now $(x-1)(x-3)$

$x-1=\frac{6+3\cos \theta }{5+4\cos \theta }-1$

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

$=\frac{1-\cos \theta }{5+4\cos \theta }$

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

$=\frac{6+3\cos \theta -15-12\cos \theta }{5+4\cos \theta }$

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

$(1-1)(x-3)=\frac{1-\cos \theta }{5+4\cos \theta }\times \frac{-9-9\cos \theta }{5+4\cos \theta }$

$=\frac{-9(1-\cos \theta )(1+\cos \theta )}{(5+4\cos \theta {)}^2}=\frac{-9(1-{\cos }^2\theta )}{(5+4\cos \theta {)}^2}.\frac{-9\sin 2\theta }{(5+4\cos \theta {)}^2}$

$(x-1)(x-3)=y^2$