Set of values of a complex number when a given expression involving the number is given to be real

Question

If $w = \alpha + i\beta$, $\beta \ne 0$ and $z \ne 1$ satisfies the condition that $\frac{w-wz}{1-z}$ is purely real, then set of values of $z$ is $$(a) |z| = 1, z \ne 2 $$ $$(b) |z| = 1, z \ne 1 $$ $$(c) z = \overline{z} $$ $$(d) None $$

My attempt: I assumed $ z = x + iy$. Substituting this in the given expression and assuming that the expression reduces to a purely real value implies that the imaginary part must be zero. Hence we reach to a simplified equation $$(x^2 - 2x + 1 + y^2) = 0 $$ $$=> (x-1)^2 + y^2 = 0$$

The right option is provided as (b). I can't conclude to the right option from here. Could you please help me to solve the question. Any help will be appreciated.

Answer 1

HINT: For $z\neq1$ your expression is equal to $w$, so not real.

And if you are still unsure, the answer is (a). The last (a). ;-)

Answer 2

If $z\neq1$, we have $\frac{w-wz}{1-z}=\frac{w(1-z)}{1-z}=w=\alpha+\beta i$. That is never purely real if $\beta\neq0$.

Answer 3

we have $$\frac{w(1-z)}{1-z}=w$$ if $z\ne 1$

Answer 4

If the expression is real

$$\frac{w-wz}{1-z}=w\frac{1-z}{1-z}=w$$

and $$w=\bar w$$

is not possible since $\beta \ne 0$.