Question

If $z=x+iy$ and $w=\frac{1-zi}{z-i},$ show that $|W|=1\Rightarrow z$ is purely real.

Answer 1

We have. $|w|=1$

$\left|\frac{1-zi}{z-i}\right|=1\Rightarrow \left|\frac{1-zi}{z-i}\right|=1[\therefore \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}]$

$\Rightarrow |1-zi|=|z-i|\Rightarrow |1-(x+iy)|=|x+iy-i|[\because z=x+iy]$

$\Rightarrow |1+y-xi|=|x+(y-1{)}^2$

$\Rightarrow \sqrt{(1+y)+x^2}=\sqrt{x^2+(y-1{)}^2}$

$\Rightarrow 1+2y+y^2+x^2+y^2-2y+1\Rightarrow y=0$

$\therefore z=x+0i=x,$ which is purely real.