A conformal mapping onto a region bounded by convex contours (Ahlfors) I want to solve the following exercise (from Ahlfors' text, page 261)

*3. Using Ex. 2, show that $p + q$ maps $\Omega$ in a one-to-one manner onto a region bounded by convex contours. Comments:

(i) A closed curve is said to be convex if it intersects every straight line at most twice.
(ii) To prove that the image of $C_k$ under $p+q$ is convex we need only show that for every $\alpha$ the function $\Re (p + q)e^{i \alpha}$ takes no value more than twice on $C_k$. But $\Re (p +q)e^{i\alpha}$ differs from $\Re (q \cos \alpha +ip \sin \alpha)$ only by a constant, and the desired conclusion follows by the properties of the mapping function in Ex. 2.
(iii) Finally, the argument principle can be used to show that the images of the contours $C_k$ have winding number $0$ with respect to all values of $p+q$. This implies, in particular, that the convex curves lie outside of each other.
As a reference, exercise 2 was 

${}$2. Show that the function $e^{-i \alpha}(q \cos \alpha +ip \sin \alpha)$ maps $\Omega$ onto a region bounded by inclined slits.

Here, $\Omega$ is a a region of finite connectivity, bounded by analytic curves $C_1,C_2, \dots,C_n$. 
$p: \Omega \to \hat{\mathbb C}$ is a complex meromorphic function, determined by


*

*Its only pole is a simple one at $z_0$, with residue 1.

*Its real part is constant on each of the contours $C_k$.


$q: \Omega \to \hat{\mathbb C}$ is similarly determined except its imaginary part is constant on the contours.

I've managed to solve Ex. 2 completely, and regarding Ex. 3 I'm stuck at step (iii).
Given that the contours $C_k \subset \partial \Omega$ are mapped onto convex contours, how can I prove that all of their images have winding number zero with respect to all values of $p+q$?
I guess that the next step is observing that for any $w_0$ taken by $p+q$ we have $$\frac{1}{2 \pi i} \sum_{k=1}^N \oint_{C_k} \frac{p'(z)+q'(z)}{p(z)+q(z)-w_0} \mathrm{d} z=N-P=\text{Ind} \left[(p+q)(\sum_{k=1}^N C_k),w_0 \right] $$
Where $N,P$ are the number of zeros,poles (respectively) of the denominator of the integrand in $\Omega$. However, I can't see how to make progress from here.
Any help is greatly appreciated!
Thanks!
EDIT:
Since my goal is determining the winding number using the argument principle, I believe I should try to evaluate $N-P$:
Well, $P=1$ since $p+q-w_0$ has a single simple pole at $z_0 \in \Omega$. And if $w_0$ is taken by $p+q$ then $N \geq 1$. I still cannot see why $N$ can't be strictly greater than $1$ - and even if it isn't, this is only the winding number of the image of the entire boundary, and not of its individual components as asked.
 A: Let $\gamma_k$ be the image of $C_k$ under $p+q$.
Since $\gamma_k$ is convex, there are only three types of points on the $w$-plane.

*

*On-curve: $w$ is on $\gamma_k$.

*External: We can draw a straight line connecting $w$ and infinity without intersecting $\gamma_k$. Obviously the winding number of $\gamma_k$ with respect to an external point is 0.

*Internal: We cannot draw a straight line connecting $w$ and infinity without intersecting $\gamma_k$.

Now let us assume that $w$ is a value of $p+q$, i.e. $p(z) + q(z) = w$ for a certain $z \in \Omega$.
$w$ cannot be internal, because this would be in contradiction with the fact that $\Omega$ is connected and $z_0 \in \Omega$ is mapped to infinity.
If $w$ is external, we are done.
If $w$ is on-curve, either it is on a slit, or it has internal points in its neighborhood.
In the first case, we can make the same argument under (17) on Page 260 in Ahlfors that the principle part of the following integral must vanish.
$$\int_{C_k} \frac{p'(z) + q'(z)}{p(z)+q(z)-w} dz$$
The second case is impossible because of the following corollary (Ahlfors, Page 132).
Corollary 1. A nonconstant analytic function maps open sets onto open sets.
