The equation of the locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$, where $z=x+i y$ is a complex number, is

Question

Aicharya · Accepted Answer

$\left|\frac{z-i}{z+i}ight|=2$ $\because \quad z=x+i y$ $	herefore \quad\left|\frac{x+i y-i}{x+i y+i}ight|=2$ $\Rightarrow \quad\left|\frac{x+(y-1) i}{x+(y+1) i}ight|=2$ $\Rightarrow \quad|x+i(y-1)|=2|x+(y+1) i|$ $\Rightarrow \quad x^2+(y-1)^2=4\left(x^2+(y+1)^2ight)$ $\Rightarrow x^2+y^2-2 y+1=4 y^2+8 y+4+4 x^2$ $\Rightarrow \quad 3 x^2+3 y^2+10 y+3=0$

The equation of the locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$, where $z=x+i y$ is a complex…

Solution