Mathematics - Complex Number Question with Solution | TestHub

Let $z=x+iy$ $\frac{z-i}{z+i}$ is purely imaginary means its real part is zero. $$\begin{array}{rl}& \frac{x+iy-i}{x+iy+i}=\frac{x+i(y-1)}{x+i(y+1)}\times \frac{x-i(y+1)}{x-i(y+1)}\\ & =\frac{x^2-2ix(y+1)+xi(y-1)+y^2-1}{x^2+(y+1{)}^2}\\ & =\frac{x^2+y^2-1}{x^2+(y+1{)}^2}-\frac{2xi}{x^2+(y+1{)}^2}\end{array}$$ for pure imaginary, we have $$\begin{array}{rl}& \frac{x^2+y^2-1}{x^2+(y+1{)}^2}=0\\ \Rightarrow & x^2+y^2=1\\ \Rightarrow & (x+iy)(x-iy)=1\\ \Rightarrow & x+iy=\frac{1}{x-iy}=z\end{array}$$ and $\frac{1}{z}=x-iy$ $$z+\frac{1}{z}=(x+iy)+(x-iy)=2x$$ $\left(z+\frac{1}{z}\right)$ is any non-zero real number