Where does $z$ lie, if $\left|\frac{z-5 i}{z+5 i}\right|=1$ ?
Let $$z=x+i y$$
Given that, $\quad\left|\frac{z-5 i}{z+5 i}\right|=\left|\frac{x+i y-5 i}{x+i y+5 i}\right|$
$$\begin{array}{ll} \Rightarrow & \left|\frac{z-5 i}{z+5 i}\right|=\frac{|x+i(y-5)|}{|x+i(y+5)|} \\ \Rightarrow & \left|\frac{z-5 i}{z+5 i}\right|=\frac{\sqrt{x^2+(y-5)^2}}{\sqrt{x^2+(y+5)^2}} \quad \left[\because\left|\frac{z-5 i}{z+5 i}\right|=1\right] \end{array}$$
On squaring both sides, we get
$$\begin{array}{rlrl} & x^2+(y-5)^2 =x^2+(y+5)^2 \\ \Rightarrow & -10 y =+10 y \\ \Rightarrow & 20 y =0 \\ \therefore & y =0 \end{array}$$
So, $z$ lies on real axis.
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