Question

please solve the 27 th​

Answer 1

$\frac{z - 1}{z + 1} \\ = \frac{x - iy}{x - iy}$

$\frac{(x - 1 + y)(x + 1 - y)}{(x + 1)^{2} + y^{2} }$

$\frac{ {x}^{2} - 1 + {y}^{2} + i(y(x + 1) - y(x - 1)) }{ {(x + 1)}^{2} + {y}^{2} }$

$now \: if \: this \: complex \: number \: is \: \\ purely \: imaginary \: then \: we \: must \: have \: \\ {x}^{2} - 1 + {y}^{2} = 0$

$or \: {x}^{2} + {y}^{2} = 1$

$or \: |z| = 1$