1. Tardigrade
  2. Question
  3. Mathematics
  4. If |z|=1,(z ≠-1) and z=x+ iy, then ((z-1/z+1)) is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $|z|=1,(z \neq-1)$ and $z=x+ iy$, then $\left(\frac{z-1}{z+1}\right)$ is

Complex Numbers and Quadratic Equations

Solution:

$z=x+i y$
$\Rightarrow |z|^{2}=x^{2}+y^{2}=1 \ldots(i)$
Now, $\left(\frac{z-1}{z+1}\right)=\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}$
$=\frac{\left(x^{2}+y^{2}-1\right)+2 i y}{(x+1)^{2}+y^{2}}=\frac{2 i y}{(x+1)^{2}+y^{2}}$
[By equation (i)]
Hence, $\left(\frac{z-1}{z+1}\right)$ is purely imaginary