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

Let $$z = a + bi$$ with $$a, b \in \mathbb{R}$$. Then $$\frac{z-1}{z+1} = \frac{(a-1)+bi}{(a+1)+bi}$$. Multiplying numerator and denominator by the conjugate of the denominator gives a real part equal to $$\frac{(a-1)(a+1)+b^{2}}{(a+1)^{2}+b^{2}} = \frac{a^{2}+b^{2}-1}{(a+1)^{2}+b^{2}}$$.

For $$\frac{z-1}{z+1}$$ to be purely imaginary, its real part must be zero: $$a^{2}+b^{2}-1 = 0$$, which means $$|z| = 1$$. Additionally, the imaginary part must be nonzero; the imaginary part works out to $$\frac{2b}{(a+1)^{2}+b^{2}}$$, so we need $$b \neq 0$$. Also $$z \neq -1$$ is required for the expression to be defined.

Now consider $$z + \frac{1}{z}$$. Since $$|z| = 1$$, we have $$\frac{1}{z} = \bar{z}$$, so $$z + \frac{1}{z} = z + \bar{z} = 2a$$. This is always a real number. Since $$|z| = 1$$ and $$b \neq 0$$, we have $$a^{2} < 1$$, meaning $$a \in (-1, 1)$$, so $$z + \frac{1}{z} = 2a$$ can take any real value in $$(-2, 2)$$. In particular, $$z + \frac{1}{z}$$ is any non-zero real number for any choice of $$z$$ with $$a \neq 0$$.