Question

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

• A. 0
• B. Any non-zero real number other than 1.
• C. Any non-zero real number.
• D. A purely imaginary number.

Answer 1

Answer: (C)

Any non-zero real number.

Let 'z' be $x+iy$
Given $\left(\frac{z-i}{z+i}\right)$ is a purely imaginary number.
$\frac{x+iy-i}{x+iy+i}=\frac{x+i(y-1)}{x+i(y+1)}$
$=\frac{(x+i(y-1\left)\right)(x-i(y+1\left)\right)}{x^2+(y+1{)}^2}$
$=\frac{x^2+(y^2-1)-ix(y+1)+ix(y-1)}{x^2+(y+1{)}^2}$ .
$=\frac{x^2+(y^2-1)-2ix}{x^2+(y+1{)}^2}$
$x^2+(y^2-1)=0$
$x^2+y^2=1$ .....(1)
Now, $z+\frac{1}{z}$
$=x+iy+\frac{1}{x+iy}$
$\Rightarrow x+iy+\frac{x-iy}{x^2+y^2}$
$=2x$ (From 1)
Any non zero real number.