The set of all α ∈ R, for which w = 1 + (1 - α)z/(1 - z) is a purely imaginary number, for all z ∈ C satisfying |z| = 1 and Re z ≠ 1, is

Let w = 1 + (1 - α)z/(1 - z). Since w is purely imaginary, its real part is 0. We have:
w = 1 + (1 - α)z/(1 - z) = (1 - z + (1 - α)z)/(1 - z) = (1 - αz)/(1 - z)
Let z = x + iy, where x and y are real numbers. Since |z| = 1, we have x² + y² = 1. Also, Re(z) ≠ 1, so x ≠ 1.
Then
w = (1 - α(x + iy))/(1 - (x + iy)) = (1 - αx - iαy)/(1 - x - iy)
Multiplying the numerator and denominator by the complex conjugate of the denominator:
w = [(1 - αx - iαy)(1 - x + iy)]/[(1 - x)² + y²] = [(1 - αx)(1 - x) + αy² + i(y(1 - αx) - (1 - x)αy)]/[(1 - x)² + y²]
The real part of w is:
Re(w) = [(1 - αx)(1 - x) + αy²]/[(1 - x)² + y²] = 0
Since the denominator is always positive, we require the numerator to be 0:
(1 - αx)(1 - x) + αy² = 0
(1 - x) - αx(1 - x) + αy² = 0
(1 - x)(1 - αx) + αy² = 0
Since x² + y² = 1, y² = 1 - x². Substituting this:
(1 - x)(1 - αx) + α(1 - x²) = 0
(1 - x)[1 - αx + α(1 + x)] = 0
(1 - x)(1 + α) = 0
This implies either 1 - x = 0 or 1 + α = 0.
If 1 - x = 0, then x = 1, which is not allowed (Re(z) ≠ 1).
Therefore, we must have 1 + α = 0, which means α = -1.
Thus, the set of all α ∈ R for which w is purely imaginary is {-1}.