Question

If $\alpha +i\beta ={\tan }^{-1}z$, $z=x+iy$ and $\alpha$ is constant then the locus of $z$ is

• A. $x^2+y^2+2x\cot 2\alpha =1$
• B. $\cot 2\alpha (x^2+y^2)=1+x$
• C. $x^2+y^2+2y\tan 2\alpha =1$
• D. $x^2+y^2+2x=1$

Answer 1

The correct option is A $x^2+y^2+2x\cot 2\alpha =1$

${\tan }^{-1}(x+iy)=\alpha +i\beta$
also, ${\tan }^{-1}(x-iy)=\alpha -i\beta$
So,
${\tan }^{-1}(x+iy)+ta{n}^{-1}(x-iy)=2\alpha$

$\Rightarrow {\tan }^{-1}\frac{(x+iy)+(x-iy)}{1-(x+iy)(x-iy)}=2\alpha$

So, $\frac{2x}{1-(x^2+y^2)}=\tan 2\alpha$

or $2x\cot 2\alpha =1-(x^2+y^2)$

or $(x^2+y^2)+2xcot2\alpha =1$