Solutions of $ 2- \frac{1}{z} = z^* $ in $\mathbb{C}$ I am asked to show that the only solution of the equation $ 2- \frac{1}{z} = z^* $ is $z=1$.
I have tried this a few ways the two where i get closest are as follows.
$\frac{2z-1}{z}=z^*$ 
$0=z^* z -2z +1$
$\left|z^2 \right| -2z +1=0 = \left|z \right|^2 -2z +1 $
id really like to say $(\left| z \right| -1)^2=0 $ or even better $(z -1)^2=0 $ 
ive tried a few other things but i always end up with $ \left| z \right| $ instead of $z$

$$\left|z^2 \right| -2z +1=0 \\ \to  \sqrt{x^2+y^2} ^2 -2(x+iy) +1=0\\
x^2+y^2-2x+1-i(2y)=0 $$ I t is better to say $$x^2+y^2-2x+1-i(2y)=0 +i0 \to \\
\begin{cases}x^2+y^2-2x+1=0 & (x-1)^2+y^2=0\\-2y=0 & \to y=0 \end{cases}\\
\begin{cases}x^2+y^2-2x+1=0 & (x-1)^2+0^2=0 \to\\-2y=0 & \to y=0 \end{cases}\\(x-1)^2=0 \to x=1$$ 
so 
$$z=x+iy=1+0y
$$
