If |z| = 1 (z neq -1) and z = x + iy,then lef | vedclass

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(B) Given $|z| = 1$,so $|z|^2 = x^2 + y^2 = 1$ .....$(i)$Now,consider the expression $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.Multipl

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Purely imaginary

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(B) Given $|z| = 1$,so $|z|^2 = x^2 + y^2 = 1$ .....$(i)$Now,consider the expression $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.Multiply the numerator and denominator by the conjugate of the denominator,$(x + 1) - iy$:$\frac{z - 1}{z + 1} = \frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2}$$= \frac{(x^2 - 1) + y^2 + i(y(x + 1) - y(x - 1))}{(x + 1)^2 + y^2}$$= \frac{(x^2 + y^2 - 1) + i(xy + y - xy + y)}{(x + 1)^2 + y^2}$$= \frac{(1 - 1) + 2iy}{(x + 1)^2 + y^2}$ [using equation $(i)$]$= \frac{2iy}{(x + 1)^2 + y^2}$Since the real part is $0$,the expression is purely imaginary.

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

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