Consider the function $f \colon \mathbb C \setminus \{0\} \rightarrow \mathbb C$ defined as the average between $z$ and $1/z$, that is $f(z)$ = $\frac12(z+1/z)$.
a) show that $f^{-1}(w)$ consists of two points for all $w \in \mathbb C$, except for two values of $w$.
My attempt:
I swapped the variables and got that: $y^2= 2w-1$.
So $y=\sqrt{2w-1}$ or $y = - \sqrt{2w-1}$.
Thus these two $y$'s consist the two points for all $w \in \mathbb C$.
But what are the exceptions on $w$? I guess I found one that is $w \neq 1/2$, but what's the other?
And I don't know how to solve part b):
b) show that $f$ sends the unit circle onto the segment $[-1,1] \subset \mathbb R$ contained in the real axis.
Any help please? Thank you.
We have $f(z)=w \iff z^2+2zw+1=0 \iff z^2-2zw+w^2=w^2-1 \iff (z-w)^2=w^2-1. $
If $w_1$ and $w_2$ are the roots of $w^2-1$, then $f(z)=w \iff z=w+w_1$ or $z=w+w_2.$
We have $w_1=w_2 \iff w= \pm 1.$
For b): if $|z|=1$, then $z=e^{it}$ with $t \in \mathbb R.$ We get $f(z)=\frac{1}{2}(e^{it}+e^{-it})=\frac{1}{2}( \cos t +i \sin t + \cos t +-i \sin t)= \cos t.$