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)

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