Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$0$

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

By multiplying the numerator and denominator with $2+\cos \theta -i\sin \theta$

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

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

$(x-3)(x-1)+y^2=\left(\frac{-9(1+\cos \theta )}{5+4\cos \theta }\right)\left(\frac{1-\cos \theta }{5+4\cos \theta }\right)+{\left(\frac{3(-\sin \theta )}{5+4\cos \theta }\right)}^2=\frac{-9+9{\cos }^2\theta +9{\sin }^2\theta }{{(5+4\cos \theta )}^2}=0$