How to prove $f(z)=z+\frac{1}{z}$ maps circles with $r\ne 1$ onto ellipses?

Question

How to prove $f(z)=z+\frac{1}{z}$ maps circles with $r> 1$ onto ellipses of the form $$\frac{x^2}{(r+\frac{1}{r})^2}+\frac{y^2}{(r-\frac{1}{r})^2}=1?$$ It is simple to understand the mapping to part. You just replace $z$ with $re^{I\theta}$ which simplifies to $$ k(\theta)=f(re^{i\theta})=re^{i\theta}+\frac{1}{re^{i\theta}}=(r+\frac{1}{r})\cos(\theta)+i(r-\frac{1}{r})\sin(\theta)$$ which is the parameterization of the ellipse given above. However, I do not understand how it maps onto. I tried to produce a right inverse and got $h(y)=-i\ln(\frac{y+\sqrt{y^2-4}}{2r})$ and $g(y)=-i\ln(\frac{y-\sqrt{y^2-4}}{2r})$ which works but is not a real number. I believe this is a problem. How would I produce a correct right inverse or prove it is onto?

Answer 1

Let $S(r)$ denote the circle with center $0$ and radius $r \ne 1$ and $E(r)$ the ellipse defined in your question. You have shown that $I(r) = f(S(r)) \subset E(r)$.

For $M \subset \mathbb{C}$ define $-M = \{ -z \mid z \in M\}$. We have $S(r) = -S(r)$ and $E(r) = -E(r)$.

We have $f(-z) = -f(z)$ which implies $f(-M) = -f(M)$ for $M \subset \mathbb{C} \setminus \{ 0 \}$. Hence $I(r) = f(S(r)) = f(-S(r)) = -f(S(r)) = -I(r)$.

Now assume that $I(r) \ne E(r)$. Choose $w \in E(r) \setminus I(r)$. We have $w \notin I(r)$, hence also $-w \notin -I(r) = I(r)$. Therefore $I(r) \subset E'(r) = E(r) \setminus \{ w,-w \}$. The set $E'(r)$ has two connected components $E_i(r)$ which satisfy $E_1(r) = -E_2(r)$.

$I(r)$ is the continuous image of the connected set $S(r)$, hence it is connected and we conclude that $I(r) \subset E_i(r)$ either for $i=1$ or $i=2$, wl.o.g. for $i=1$. But then also $I(r) = -I(r) \subset - E_1(r) = E_2(r)$ which is impossible.

Therefore we must have $I(r) = E(r)$.

Answer 2

Let $z=x+iy$ with $x^2+y^2=r^2$.

We have

$$z+\frac1z=x+iy+\frac{x-iy}{x^2+y^2}=x\left(1+\frac1{r^2}\right)+iy\left(1-\frac1{r^2}\right)=u+iv.$$

From this,

$$\left(\frac{u}{1+\dfrac1{r^2}}\right)^2+\left(\frac{v}{1-\dfrac1{r^2}}\right)^2=r^2.$$

Note that the transformation is a mere scaling.