Complex Number Given Magnitude and Formula

Question

Let $w$ and $z$ be complex numbers such that $w=\frac{1}{1-z}$, and $|z|^2=1$. Find the real part of $w$.

The answer is $\frac{1}{2}$ but I don't know how to get to it.

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My attempt

as $|z|^2=1$

$z\bar z = 1$

If $z = x+yi$

$z=\frac{1}{\bar z} = \frac{1}{x-yi}$

$\therefore w= \frac{1}{1-\frac{1}{x-yi}}=\frac{x-yi}{(x-1)-yi}$

$=\frac{(x-yi)((x-1)+yi)}{((x-1)-yi)((x-1)+yi)}$

$=\frac{x^2-x+yi+y^2}{x^2-x+1+y^2}$

$=\frac{x^2-x+y^2}{x^2-x+1+y^2}+\frac{y}{x^2-x+1+y^2}i$

Hence $Re(w)=\frac{x^2-x+y^2}{x^2-x+1+y^2}$

$=\frac{x^2+y^2-x}{x^2+y^2-x+1}$

$=\frac{1-x}{1-x+1}$

$=\frac{1-x}{2-x}$

$=\frac{-1}{2-x}+\frac{2-x}{2-x}$

$=1-\frac{1}{2-x}$

$=1+\frac{1}{x-2}$ ?????

Answer 1

Note that $$w+w^* = \frac{1}{1-z} + \frac{1}{1-z^*}=\frac{2-z^*-z}{1-z^*-z+|z|^2}=1.$$

Answer 2

Pure geometric solution.

All points $z$ satisfying $|z|=1$ are on circle $\mathcal{C}$ with radius $r=1$ and center at $0$.

Now transformation $z\mapsto 1-z$ is reflection across $0$ followed by translation for $1$. So $\mathcal{C}$ ''moves'' to the right for $1$ and let this new circle be $\mathcal{C}'$.

Transformation $z\mapsto {1\over z}$ is inversion with center at $0$ with radius $1$ (i.e. circle $\mathcal{C}$). Since $0\in \mathcal{C}'$ and $\mathcal{C}'$ touches imaginary axis, $\mathcal{C}'$ maps to line $\ell$ which is parallel to imaginary axis. Finally, since points of intersection of $\mathcal{C}$ and $\mathcal{C}'$, which have clearly real part ${1\over 2}$, are invariant, the line $\ell$ has equation $z={1\over 2}$ and thus conclusion.