Let $\mathrm{z}\ne -\mathrm{i}$ be any complex number such that $\frac{\mathrm{z}-\mathrm{i}}{\mathrm{z}+\mathrm{i}}$ is a purely imaginary number then $\mathrm{z}+\frac{1}{\mathrm{z}}$ is

$\text{ Let 'z' be }x+\text{ iy }$

$\text{ Given }\left(\frac{z-i}{z+i}\right)\text{ is a purely imaginary number. }$

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

$=\frac{(\mathrm{x}+\mathrm{i}(\mathrm{y}-1\left)\right)(\mathrm{x}-\mathrm{i}(\mathrm{y}+1\left)\right)}{{\mathrm{x}}^2+(\mathrm{y}+1{)}^2}$

$=\frac{x^2+\left(y^2-1\right)-ix(y+1)+ix(y-1)}{x^2+(y+1{)}^2}$

$=\frac{x^2+\left(y^2-1\right)-2ix}{x^2+(y+1{)}^2}$

$x^2+\left(y^2-1\right)=0 \Rightarrow x^2+y^2=1$

$,\mathrm{z}+\frac{1}{\mathrm{Z}}=x+iy+\frac{1}{x+iy}$

$\Rightarrow x+iy+\frac{x-iy}{x^2+y^2}=x+iy+x-iy=2x$