If frac{z - i}{z + i} (z ne -i) is a purely i | vedclass

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(B) Let $z = x + iy$. Then $\frac{z - i}{z + i} = \frac{x + i(y - 1)}{x + i(y + 1)}$.To simplify,multiply the numerator and denominator by the conjuga

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(B) Let $z = x + iy$. Then $\frac{z - i}{z + i} = \frac{x + i(y - 1)}{x + i(y + 1)}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator: $\frac{x + i(y - 1)}{x + i(y + 1)} 	imes \frac{x - i(y + 1)}{x - i(y + 1)}$.$= \frac{x^2 - ix(y + 1) + ix(y - 1) + (y - 1)(y + 1)}{x^2 + (y + 1)^2} = \frac{(x^2 + y^2 - 1) - 2ix}{x^2 + (y + 1)^2}$.Since the expression is purely imaginary,the real part must be zero:$x^2 + y^2 - 1 = 0 \implies x^2 + y^2 = 1$.Since $z \cdot \bar{z} = |z|^2 = x^2 + y^2$,we have $z \cdot \bar{z} = 1$.

If $\frac{z - i}{z + i}$ $(z \ne -i)$ is a purely imaginary number,then $z \cdot \bar{z}$ is equal to

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