Let z ≠-i be any complex number such that z- i z+ i is a purely imaginary number. Then z+ 1 z is:

Let $z=x+i y$z=x+iy $\frac{z-i}{z+i}$z+iz−i​ is purely imaginary means its real part is zero. $$\begin{aligned} &\frac{x+i y-i}{x+i y+i}=\frac{x+i(y-1)}{x+i(y+1)} \times \frac{x-i(y+1)}{x-i(y+1)} \\ &=\frac{x^2-2 i x(y+1)+x i(y-1)+y^2-1}{x^2+(y+1)^2} \\ &=\frac{x^2+y^2-1}{x^2+(y+1)^2}-\frac{2 x i}{x^2+(y+1)^2} \end{aligned}$$​x+iy+ix+iy−i​=x+i(y+1)x+i(y−1)​×x−i(y+1)x−i(y+1)​=x2+(y+1)2x2−2ix(y+1)+xi(y−1)+y2−1​=x2+(y+1)2x2+y2−1​−x2+(y+1)22xi​​ for pure imaginary, we have $$\begin{aligned} & \frac{x^2+y^2-1}{x^2+(y+1)^2}=0 \\ \Rightarrow & x^2+y^2=1 \\ \Rightarrow &(x+i y)(x-i y)=1 \\ \Rightarrow & x+i y=\frac{1}{x-i y}=z \end{aligned}$$⇒⇒⇒​x2+(y+1)2x2+y2−1​=0x2+y2=1(x+iy)(x−iy)=1x+iy=x−iy1​=z​ and $\frac{1}{z}=x-i y$z1​=x−iy $$z+\frac{1}{z}=(x+i y)+(x-i y)=2 x$$z+z1​=(x+iy)+(x−iy)=2x $\left(z+\frac{1}{z}\right)$(z+z1​) is any non-zero real number