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(C) Let $z = x + iy$. Given that $\frac{z - i}{z + i}$ is purely imaginary,its real part must be zero. $\frac{z - i}{z + i} = \frac{x + i(y - 1)}{x +

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any non-zero real number.

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(C) Let $z = x + iy$. Given that $\frac{z - i}{z + i}$ is purely imaginary,its real part must be zero. $\frac{z - i}{z + i} = \frac{x + i(y - 1)}{x + i(y + 1)} 	imes \frac{x - i(y + 1)}{x - i(y + 1)} = \frac{x^2 + i(x(y - 1) - x(y + 1)) + (y - 1)(y + 1)}{x^2 + (y + 1)^2} = \frac{x^2 + y^2 - 1 - 2xi}{x^2 + (y + 1)^2}$. For the expression to be purely imaginary,the real part must be zero: $\frac{x^2 + y^2 - 1}{x^2 + (y + 1)^2} = 0 \Rightarrow x^2 + y^2 = 1$. Since $|z|^2 = x^2 + y^2 = 1$,we have $z \bar{z} = 1$,which implies $\bar{z} = \frac{1}{z}$. Then $z + \frac{1}{z} = z + \bar{z} = (x + iy) + (x - iy) = 2x$. Since $z 
eq -i$,$x^2 + y^2 = 1$ implies $z$ lies on the unit circle excluding the point $(0, -1)$. Thus,$x$ can take any real value in the interval $(-1, 1)$,but $x$ cannot be $0$ because if $x=0$,then $y^2=1$,so $y=1$ (as $y=-1$ is excluded). Thus $2x$ can be any real number except $0$.

Let $z \neq -i$ be any complex number such that $\frac{z - i}{z + i}$ is a purely imaginary number. Then $z + \frac{1}{z}$ is

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