If z is a complex number such that frac{z - 1 | vedclass

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(B) Let $w = \frac{z - 1}{z + 1}$. Since $w$ is purely imaginary,$w + \overline{w} = 0$.Substituting $w = \frac{z - 1}{z + 1}$,we get $\frac{z - 1}{z

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$|z| = 1$

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(B) Let $w = \frac{z - 1}{z + 1}$. Since $w$ is purely imaginary,$w + \overline{w} = 0$.Substituting $w = \frac{z - 1}{z + 1}$,we get $\frac{z - 1}{z + 1} + \overline{\left(\frac{z - 1}{z + 1}\right)} = 0$.$\frac{z - 1}{z + 1} + \frac{\overline{z} - 1}{\overline{z} + 1} = 0$.$(z - 1)(\overline{z} + 1) + (\overline{z} - 1)(z + 1) = 0$.$(z\overline{z} + z - \overline{z} - 1) + (z\overline{z} + \overline{z} - z - 1) = 0$.$2z\overline{z} - 2 = 0$.$z\overline{z} = 1$.Since $|z|^2 = z\overline{z}$,we have $|z|^2 = 1$,which implies $|z| = 1$.

If $z$ is a complex number such that $\frac{z - 1}{z + 1}$ is purely imaginary,then

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