Let z-i be any complex number such that z-iz+i is a purely imaginary number.

Question

Let $z
e-i$ be any complex number such that $\dfrac{z-i}{z+i}$ is a purely imaginary number. Then $z+\dfrac1z$ is:

Accepted Answer

Purely imaginary condition gives $x^2+y^2=1$ (with $z=x+iy$), so $\dfrac1z=x-iy=\bar z$.

$$z+\frac1z=(x+iy)+(x-iy)=2x,$$

any non-zero real number as $x$ ranges over $(-1,1)\setminus\{0\}$.

Solutions · 1