Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

Consider the given equation

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

And find the value $(x-1)(x-3)+y^2=?$

On rationalizing and we get,

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

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

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

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

On comparing and we get,

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

Now,

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

$=(\frac{6+3\cos \theta }{5+4\cos \theta }-3)(\frac{6+3\cos \theta }{5+4\cos \theta }-1)+{\left(\frac{-3sin\theta }{5+4\cos \theta }\right)}^2$

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

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

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

$=0$

Hence this is the answer.