Question

Show that if $\left|\frac{z-3i}{z+3i}\right|=1,$ then $z$ is a real number.

Answer 1

$|\left(\frac{z-3i}{z+3i}\right)|=1|z-3i|=|z+3i||x+iy-3i|=|x+iy+3i||x+i(y-3)|=|x+i(y+3)|\sqrt{x^2+(y-3{)}^2}=\sqrt{x^2+(y+3{)}^2}{x}^2+(y-3{)}^2=x^2+(y+3{)}^2(y-3{)}^2=(y+3{)}^{2}y-3=\pm (y+3)\therefore y-3=y+3or-3=3whichisnotpossibletheny-3=-(y+3)ory=0soz=x+iy=x+0=x$

z is real number.