Let z ≠ -i be any complex number such that z - iz + i is a purely imaginary number. Then z + 1/z is:

Let 'z' be x + iy
Given (z - iz + i) is a purely imaginary number.
x + iy - i(x + iy) + i = x + iy - ix - i²y + i = x + iy - ix + y + i = x + y + i(y - x + 1)
Since it is purely imaginary, the real part must be zero.
x + y = 0 => y = -x
Now, let's find z + 1/z
z + 1/z = x + iy + 1/(x + iy) = x + iy + (x - iy)/(x² + y²)
Since y = -x
z + 1/z = x - ix + 1/(x - ix) = x - ix + (x + ix)/(x² + x²) = x - ix + (x + ix)/(2x²)
= x - ix + (x + ix)/(2x²) = x - ix + 1/(2x) + i/(2x) = x + 1/(2x) + i(-x + 1/(2x))
Since the imaginary part is non-zero unless x = 1/2, and the real part is non-zero, this expression will be a non-zero real number only if the imaginary part is zero, which is only true if x=1/(2x), so x² = 1/2 which implies x = ±1/√2. Thus, the imaginary part is zero only when x = 1/2 which leads to a real part equal to 1.
If x + y = 0, then y = -x.
z + 1/z = x + iy + x - iy / (x² + y²) = x + iy + (x - iy)/(2x²) = x - ix + (x + ix)/(2x²) = x - ix + 1/(2x) + i/(2x) = x + 1/(2x) + i(1/(2x) - x) = 1 + 1/1-i = 1 + i/2
If x + y = 0, z = x - ix
z + 1/z = x - ix + 1/(x - ix) = x - ix + (x + ix)/(2x²) = x - ix + 1/(2x) + i/(2x) = x + 1/(2x) + i(1/(2x) - x)
For this to be real, 1/(2x) - x = 0 => 1 = 2x² => x = ±1/√2
Then z + 1/z = x + 1/(2x) = 1/√2 + √2/2 = √2
Therefore, z + 1/z is any non-zero real number.