If frac{z-1}{z+1} is purely imaginary, then | vedclass

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(B) Let $\omega = \frac{z-1}{z+1}$ be a purely imaginary number. $A$ complex number $\omega$ is purely imaginary if and only if $\omega + \overline{\o

Vedclass · Accepted Answer

$|z|=1$

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(B) Let $\omega = \frac{z-1}{z+1}$ be a purely imaginary number. $A$ complex number $\omega$ is purely imaginary if and only if $\omega + \overline{\omega} = 0$ (where $\omega 
eq 0$). So, $\frac{z-1}{z+1} + \overline{\left(\frac{z-1}{z+1}ight)} = 0$ $\frac{z-1}{z+1} + \frac{\overline{z}-1}{\overline{z}+1} = 0$ $(z-1)(\overline{z}+1) + (\overline{z}-1)(z+1) = 0$ $(z\overline{z} + z - \overline{z} - 1) + (z\overline{z} - z + \overline{z} - 1) = 0$ $2z\overline{z} - 2 = 0$ $z\overline{z} = 1$ Since $|z|^2 = z\overline{z}$, we have $|z|^2 = 1$, which implies $|z| = 1$.

If $\frac{z-1}{z+1}$ is purely imaginary, then

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