The transformation $z=w+1 / w$ maps the circle $|w|=1$ onto the line segment joining the points $z=2$ and $z=-2$, and it maps the domain outside that circle onto the rest of the $z$ plane. (See Exercises 18 and 19, Sec. 41.) Write
$$
z-2=r_1 \exp \left(i \theta_1\right), \quad z+2=r_2 \exp \left(i \theta_2\right)
$$
and
$$
\left(z^2-4\right)^{1 / 2}=\sqrt{r_1 r_2} \exp \frac{i\left(\theta_1+\theta_2\right)}{2} \quad\left(0 \leqq \theta_1<2 \pi, 0 \leqq \theta_2<2 \pi\right) ;
$$
the function $\left(z^2-4\right)^{1 / 2}$ is then single-valued and analytic everywhere except on the branch cut consisting of the segment of the $x$ axis between the points $z= \pm 2$. Show that the inverse of the transformation $z=w+1 / w$, such that $|w|>1$ for every point $z$ not on the branch cut, can be written
$$
w=\frac{1}{2}\left[z+\left(z^2-4\right)^{1 / 2}\right]=\frac{1}{4}\left(\sqrt{r_1} \exp \frac{i \theta_4}{2}+\sqrt{r_2} \exp \frac{i \theta_2}{2}\right)^2 .
$$
Thus the transformation and that inverse establish a one to one correspondence between points in the two domains.