Image of open donut under $\phi=z+\frac{1}{z}$ 
Source: Ch1 problem 11 on Function Theory of One Complex Variable by Greene and Krantz.
Let $S=\{z\in C: \frac{1}{2}<|z|<2\}$ and $\phi(z)=z+\frac{1}{z}$. Compute $\phi(S)$.

So far I have $$z=re^{i\theta}, \frac{1}{2}<r<2, 0\leq\theta<2\pi\implies \phi(z)=\left(r+\frac{1}{r}\right)\cos(\theta)+i\left(r-\frac{1}{r}\right)\sin(\theta)=Re^{i\varphi}$$
$$R=\sqrt{r^2+\frac{1}{r^2}+2\cos(2\theta)}$$
$$\varphi=\arctan\left(\frac{r^2-1}{r^2+1}\tan(\theta)\right)$$
but it doesn't look very helpful...
 A: To understand the mapping behaviour of a holomorphic function $f$, it is often helpful to see how it transforms a family of curves - the most obvious candidates for such families are the lines of constant real or imaginary part, or a family of concentric circles and the rays emanating from the centre - and the preimages of such families, that is, the level curves of $\operatorname{Re} f,\; \operatorname{Im} f$ and of $\lvert f(z) - w_0\rvert$, $\arg\bigl(f(z) - w_0\bigr)$ respectively.
Another tool that often helps greatly is symmetry. If $f$ is even or odd, that helps understanding its behaviour, also if $f$ is real in the sense $f(\overline{z}) = \overline{f(z)}$. Other simple transformations are also something to look for.
Here, we have the symmetry $\phi(z) = \phi\bigl(\frac{1}{z}\bigr)$. Thus, to understand $\phi$, we only need to look at $1 \leqslant \lvert z\rvert$, the behaviour inside the unit disk follows from the behaviour outside it.
Since the domain $S$ we're interested in is an annulus centred at $0$, it is natural to look how $\phi$ behaves on the circles $\lvert z\rvert = r$, and on the rays $\arg z = \theta$.
For $r = 1$, we have $\frac{1}{z} = \overline{z}$, and hence $\phi(z) = 2\operatorname{Re} z$ on the unit circle. Thus the unit circle is mapped to the interval $[-2,2]$. Generally, you found that
$$\phi(re^{i\theta}) = \bigl(r + \tfrac{1}{r}\bigr)\cos \theta + i\bigl(r -\tfrac{1}{r}\bigr)\sin \theta.$$
For constant $r \neq 1$, that describes an ellipse with major semiaxis $r+\frac{1}{r}$ and minor semiaxis $\bigl\lvert r-\frac{1}{r}\bigr\rvert$, which is traversed counterclockwise for $r > 1$, and clockwise for $r < 1$.
To determine the image $\phi(S)$, that is already sufficient. Using the symmetry $\phi\bigl(\frac{1}{z}\bigr) = \phi(z)$, and the open mapping theorem, it then follows that $\phi(S)$ is a region bounded by the ellipse we get for $r = 2$, and since $\phi(\pm i) = 0$, it is the interior of that curve. We can write that in the form
$$\phi(S) = \{ z \in \mathbb{C} : \lvert z-2\rvert + \lvert z+2\rvert < 5\},$$
using that the foci of the ellipse $\{\phi(2e^{i\theta}) : 0 \leqslant \theta \leqslant 2\pi\}$ are $\pm 2$.
To more completely understand the mapping behaviour of $\phi$, one would also look at the curves $\bigl\{ \phi(re^{i\theta}) : \frac{1}{2} < r < 2\bigr\}$ for fixed $\theta$, and note that - for $\theta \neq k\frac{\pi}{2}$ - these are parts of the hyperbolas with foci $\pm 2$.
