Let $z=x+i y$

$\frac{z-i}{z+i}$ is purely imaginary means its real part is zero.

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

${ = \frac{{{x^2} – 2ix(y + 1) + xi(y – 1) + {y^2} – 1}}{{{x^2} + {{(y + 1)}^2}}}}$

${ =  – \frac{{{x^2} + {y^2} – 1}}{{{x^2} + {{(y + 1)}^2}}}\,\frac{{2xi}}{{{x^2} + {{(y + 1)}^2}}}}$

for pure imaginary, we have

${\frac{{{x^2} + {y^2} – 1}}{{{x^2} + {{(y + 1)}^2}}} = 0}$

$\Rightarrow x^{2}+y^{2}=1$

$\Rightarrow(x+i y)(x-i y)=1$

$\Rightarrow x+i y=\frac{1}{x-i y}=z$

and $\frac{1}{z}=x-i y$

$z+\frac{1}{z}=(x+i y)+(x-i y)=2 x$

$\left(z+\frac{1}{z}\right)$ is any non-zero real number