Question

If $\alpha +i\beta =ta{n}^{-1}\left(z\right),z=x+iy$ and $\alpha$ is constant then the locus of $z$ is:

• A. $\left(tan2\alpha \right)(x^2+y^2)=tan2\alpha -2x$
• B. $\left(cot2\alpha \right)(x^2+y^2)=1+x$
• C. $x^2+y^2+2ycot2\alpha =1$
• D. $x^2+y^2+2xsin2\alpha =1$

Answer 1

The correct option is A $\left(tan2\alpha \right)(x^2+y^2)=tan2\alpha -2x$

Given $z=x+iy$
$\Rightarrow \alpha +i\beta =ta{n}^{-1}(x+iy)$
also,
$\Rightarrow \alpha -i\beta =ta{n}^{-1}(x-iy)$
$2\alpha =ta{n}^{-1}(x-iy)+ta{n}^{-1}(x-iy)$
$2\alpha =ta{n}^{-1}\frac{(x+iy)+(x-iy)}{1-(x+iy)(x-iy)}$
$2\alpha =ta{n}^{-1}\frac{2x}{1-(x^2+y^2)}$
$\Rightarrow \left(tan2\alpha \right)(x^2+y^2)=tan2\alpha -2x$