If $\frac{z - 1}{z + 1}$ is purely imaginary number $(z
eq - 1)$, then $|z|$ is equal to

Question

If $\frac{z - 1}{z + 1}$ is purely imaginary number $(z 
eq - 1)$, then $|z|$ is equal to

Dr. Abubacker · Accepted Answer

(a)
Since, $\frac{z - 1}{z + 1}$ is purely imaginary
$	herefore\ \ \ \frac{z - 1}{z + 1} = - \overline{\left( \frac{z - 1}{z + 1} ight)}$
$\Rightarrow \ \ \ \ \frac{z - 1}{z + 1} = \frac{\overline{z} - 1}{\overline{z} + 1}$
$\Rightarrow \ \ \ \ \frac{2z}{- 2} = \frac{2}{- 2\overline{z}}\ \ \ \  \Rightarrow \ \ \ \ z\overline{z} = 1$
$\Rightarrow \ \ \ \ |z|^{2} = 1\ \ \ \  \Rightarrow \ \ \ \ |z| = 1$

Q. If $\frac{z - 1}{z + 1}$ is purely imaginary number $(z \neq - 1)$, then $|z|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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