I want to show that $f(z)=\frac{1}{2}\left(z+\frac{1}{z}\right)$ is a conformal map from the set of $z$ such that $0<|z|<1$ onto $\mathbb{C} \setminus [-1,1]$.
I find that f'(z)=\frac{(z+1)(z-1)}{2z^2} so this means that $f$ is conformal except for $z=1$ or $z=-1$ which is ok since they're not in the domain of $f$. So $f$ is a conformal mapping.
Now for $z=re^{i\theta}$, we find $w=\frac{1}{2}\left(z+\frac{1}{z}\right)=re^{i\theta}+\frac{e^{-i\theta}}{r}$ which gives $u=(r+\frac{1}{r})\frac{cos(\theta)}{2}$ and $v=(r-\frac{1}{r})\frac{sin(\theta)}{2}$. When we compute we end up with $\frac{u^2}{(r+1/r)^2} + \frac{v^2}{(r-1/r)^2}=\frac{1}{4}$ which means that circles about the origin, i.e the ones such that $|z|<1$ are mapped to ellipses.
Finally if $[-1,1]$ were in the image then this means that the unit circle has been mapped by $f$ since would take $r=1$ which implies $-1 ($v=0$ here).
Is the exercise complete or do I need to add more justification as to why $f$ maps the punctured interior of the unit disc to the whole plane?
EDIT: I corrected the typo in the equation of the ellipse, it should be $\frac{1}{4}$ like in Zarrax's answer