Question

Let $z=x+$ iy and $v=\frac{1-iz}{z-i},$ show that if $|v|=1,$ then $z$ is purely real.

Answer 1

$v=\left(\frac{1-iz}{z-i}\right)\times \left(\frac{i}{i}\right)v=\left(\frac{z+i}{z-i}\right)\times \left(\frac{1}{i}\right)\therefore |v|=\left(\frac{|z+i|}{|z-i|}\right)\times \left(\frac{1}{|i|}\right)1=\left(\frac{(x+iy+i)}{|x+iy-i|}\right)\times \left(\frac{1}{i}\right)|x+i(y-1)|=|x+i(y+1)|\sqrt{x^2+(y-1{)}^2}=\sqrt{x^2+(y+1{)}^2}{x}^2+(y-1{)}^2=x^2+(y+1{)}^2(y-1{)}^2=(y+1{)}^2\therefore y-1=\pm (y+1)ory-1=+y+1then-1=1\left(notpossible\right)theny-1=-(y+1)ory=0\therefore x=x+iy=x+0=xsozisarealnumber.$