Question

If $|\frac{z-5i}{z+5i}|=1$, show that z is a real number.

Answer 1

$\frac{z-5i}{z+5i}=1$

$\Rightarrow \frac{x+iy-5i}{x+iy+5i}=1$

$\Rightarrow \frac{x+i(y-5)}{x+i(y+5)}=1$

$\Rightarrow |\frac{x+i(y-5)}{x+i(y+5)}\times \frac{x-i(y+5)}{x-i(y+5)}|=1$

$\Rightarrow \frac{x^2+i(y-5)x-ix(y+5)+y^2-25}{x^2+{(y+5)}^2}=1$

$\Rightarrow x^2+i(yx-5x-xy-5x)+y^2-25=x^2+y^2+10y+25$

$\Rightarrow +i(-5x-5x)-25=10y+25$

$\Rightarrow -10ix-10y=50$

$\Rightarrow -i(x+iy)=5$

$\Rightarrow x+iy=\frac{5}{-i}\times \frac{i}{i}=5i$

Comparing the real and imaginary parts we get

$\therefore x=0,y=5$

$\therefore z=0+5i=5i$