A complex number is real proof

Question

The question is whether this complex number is real: $A= \frac{z-uz*}{1-u} / z \neq z*, |u| = 1.$ For $A$ to be real,$ A* = A$ is necessary. I found $A* = - A$, which says that $A \in iIR$. But, the solution says it s real. I used this as part of my proof: $|u|^2 = uu*$ (* to denote the complex conjugate) This is what I did: $A*= \frac{z*-u*z}{1-u*} \rightarrow A*= \frac{z*-\frac{1}{u}z}{1-\frac{1}{u}} \rightarrow A*= \frac{\frac{uz*-z}{u}}{\frac{u-1}{u}} \rightarrow A*= \frac{uz*-z}{u-1} = -A$. We have $|u|=1 \rightarrow u* = \frac{1}{u}$

Answer 1

Let's see.

$A=\dfrac{z-u\bar z}{1-u}$. (I prefer to write $\bar z$ for the conjugate of $z$.)

So $\bar A=\overline {\dfrac{z-u\bar z}{\overline {1-u}}}=\dfrac{\bar z-\bar u z}{1-\bar u}=\dfrac{\bar z-\bar u z}{1-\bar u}\cdot\dfrac uu=\dfrac{u\bar z-z}{u-1}=A$.

Thus $A$ is real.

Answer 2

$A* = \frac{z*-(uz*)*}{1-u*} = \frac{z*-(\frac{1}{u*}z*)*}{1-\frac{1}{u}} =\frac{z* - \frac{z}{u}}{1-\frac{1}{u}} = \frac{z - z*u}{1-u} = A$

$u* = |u|^2/u = 1/u$