Rational function level curve Consider a rational function $h$ with poles at infinity, and finitely other points.
I would like to show the following:
Let $\lambda$ be some positive number for which the level set $|h| = \lambda$ is a disjoint union of circles.
The circles may be nested in one another. Let us suppose there are several (possibly more than one) of the circles of the level set contained in the interior of some larger circle in the level set, and these are the only such circles contained in this larger circles interior.
I would like to show that we cannot have all the disks bounded by the circles contained in the  interior of the larger circle be such that $h$ maps each such disk biholomorphically onto the disk center zero of radius $\lambda$ (and so that $h$ extends continuously to the boundary and remains one to one).
For more context, I am trying to show this cannot be possible due to a remark made in a paper by Robinson, where he seems to imply that the disjoint union of the above mentioned circles wind around every point of the sub level set $|h|^{-1}[0,\lambda)$ once positively.
In trying to prove this I  seem to have reduced Robinsons claim to the content of this post.
Thank you in advance.
Edit: For full disclosure, what I reduced the claim down to was the same as what I wrote above, with "circles" replaced by "analytic Jordan curves". I thought the general idea of a proof for one should probably be the same as forthe other which is why I used circles instead.
 A: I think I managed to sort it out, although I'm not particularly thrilled with the way I went about (possibly) resolving it, so I won't accept this answer yet in the event someone chimes in with a neater solution.
So let's just assume that $|h|^{-1}(\lambda)$ is a disjoint union of analytic Jordan curves, and address the situation I posed above (namely show that such a situation cannot arise). We assume that $h$ is nonconstant, since it does not have poles besides $\infty$ and finitely other points.
Now in the situation I mention, let $\gamma_1$ denote the 'larger circle' , and $\gamma_2,...,\gamma_s$ denote the 'smaller circles' contained in the interior of $\gamma_1$, so that $h$ maps $\text{int}{\gamma_j}$ biholomorphically onto $\Delta(0,\lambda)$ for $j=2...s$, and extends continuously to the boundary so that it remains one-one.
Then claim that $\Omega = \text{int}(\gamma_1) \cap \bigcap_{j=2}^s \text{ext}(\gamma_j)$ contains a pole. Indeed if not then neither does $\text{int}(\gamma_1)$, which is absurd since we then have a holomorphic function that has constant modulus $\lambda$ on $\gamma_1$, the boundary of $\text{int}(\gamma_1)$, but obtains values of modulus $\lambda$ in $\text{int}(\gamma_1)$ on the curves $\gamma_j$, $j=2...s$.
In fact one must have that $|h| \geq \lambda$ on $\Omega$, otherwise there would be no path from some point $p \in \Omega$ where $|h| < \lambda$ to the pole in $\Omega$ as such a path would have to cross some $\lambda$-level curve of $|h|$. But $\Omega$ is manifestly a region, i.e. it is manifestly connected.
Now in the event of this, we have two possible scenarios. Either $\gamma_1$ is not contained in the interior of any other Jordan curve in the level set $|h|^{-1}(\lambda)$, or there is some innermost Jordan curve in the level set $|h|^{-1}(\lambda)$ containing it in its interior.
In the former situation, by the open mapping theorem there are points arbitrarily close to $\gamma_1$ exterior to it, such that $|h| < \lambda$ (since $|h| \geq \lambda$ in $\Omega$), but then there is no path exterior to all the Jordan curves comprising $|h|^{-1}(\lambda)$ from such a point to $\infty$, and we may select such a point to be exterior to all the Jordan curves comprising $|h|^{-1}(\lambda)$ since we assume $\gamma_1$ is not interior to any other Jordan curve. This violates the path-connectivity of the region exterior to all of the Jordan curves comprising $|h|^{-1}(\lambda)$.
In the latter situation, if $\gamma'$ is the innermost Jordan curve containing $\gamma_1$ in its interior, then denoting by $\Omega'$ the region bounded by $\gamma'$ , $\gamma_1$, and all the outermost Jordan curves contained in the interior of $\gamma'$ besides $\gamma_1$, say for definiteness these are denoted by $\gamma'_1,...,\gamma'_t$, (if $t =0$ then there are no such Jordan curves which is also possible), $\Omega'$ contains no pole of $h$ by a path-connectivity argument (if it did, there would be a path in $\Omega'$ from some point less than $\lambda$ to the pole, which would necessarily hit the boundary of $\Omega'$ which is absurd since $\Omega'$ is connected).
By the minimum modulus principle, $\Omega'$ contains some zero of $h$ ($h$ achieves absolute value less than $\lambda$ in points of $\Omega'$, is holomorphic all throughout $\Omega'$ since there is no pole of $h$ in $\Omega'$, and $\Omega'$ is bounded by $\lambda$-level curves of $|h|$).
Now we have that $$\frac{1}{2 \pi i} \int_{\partial \Omega'} \frac{h'(z)}{h(z)} dz = \eta(h \circ \partial \Omega', 0) = \eta(C_{\lambda}(0),0) - \eta(C_{\lambda}(0),0) - t \eta(C_{\lambda}(0),0) = -t  $$ $$= Z_{\Omega'} - P_{\Omega'}$$,
but $P_{\Omega'}$ is necessarily zero as we already have shown, hence $-t = Z_{\Omega'} \leq 0$ which means that $\Omega'$ has no zero, but this is absurd since we already know $\Omega'$ has a zero by the minimum modulus principle.
Thus both situations yield contradictions, which means that the situation I originally proposed cannot occur.
