Polynomial Image of a Circle In the "streamlined" proof of the FToA given here an n'th degree polynomial over $\mathbb{C}$ with complex coefficients is given, & it is said that as we choose arguments $z$ lying on a circle (with a large enough radius) in the plane the polynomial will do n loops around the origin:

First off, I don't see why this should be so, as in I don't see why one can expect any more than one loop, & second I don't see how one can be sure the loops simply must go around the origin. 
I would prefer to see this with as little assumptions as possible, as in to see the basic logic of it as the article assumes, no Morera's theorem or something like that - thank you!
 A: First take a look at the polynomial $z^n$ and convince yourself that if you choose arguments on any circle then $z^n$ will do $n$ loops around the origin (use $z = re^{\theta i}$ to see that).
Now if you have a polynomial of the form $z^n + c_{n - 1}z^{n - 1} + \cdots + c_1z + c_0$ the term $z^n$ will go around the origin $n$ times but the terms $c_{n - 1}z^{n - 1} + \cdots + c_1z + c_0$ might deform this so that it's no longer a loop around the origin.  This is where choosing a "big enough" circle comes in.  You have to choose the radius $r$ so that for any point $z = re^{\theta i}$ on the circle $|z^n| > |c_{n - 1}z^{n - 1} + \cdots + c_1z + c_0|$.  This will then make it so the "fudge factor" $c_{n - 1}z^{n - 1} + \cdots + c_1z + c_0$ can't pull the circle $z^n$ back across the origin.  The whole polynomial $z^n + c_{n - 1}z^{n - 1} + \cdots + c_1z + c_0$ may not be a circle any more but it will still loop around the origin $n$ times.
Finally note that multiplying through by a constant just makes the loops bigger, so the fact that I've chosen a monic polynomial above doesn't matter.
Last but not least, note I haven't told you how to choose the radius $r$ in that second paragraph, so this isn't a rigorous proof, it's just the idea.  You can say explicitly how to choose $r$ and make this argument rigorous but I'm not sure it adds anything to your understanding for me to do that for you.  You should first try and see that this idea is right and then try and figure out how to choose $r$ on your own.
A: If $f(z) = z^n + c_{n-1}z^{n-1} + \cdots + c_0$ and $z = re^{i\theta}$, then for $r$ large, $f(z)$ is essentially $r^n e^{in\theta}$ (I'll get back to "essentially"). This means that when we let $z$ travel along a circle of radius $r$, $f(z)$ will trace out a curve that is almost an $n$-fold circle of radius $r^n$.
The key point to make this rigorous is that if $r$ is large enough and $z=re^{i\theta}$, then
$$|c_{n-1}z^{n-1} + \cdots + c_0| \le \frac12 r^n.$$
You can show this either with the help of the triangle inequality or a limit argument. Try to provide the details yourself! Hence
\begin{align}
|f(z)| &= |z^n + c_{n-1}z^{n-1} + \cdots + c_0| \\
& \ge |z^n| - |c_{n-1}z^{n-1} + \cdots + c_0| \ge r^n - \frac12 r^n = \frac 12r^n.
\end{align}
This means that the point $f(z)$ is "close" to the point $z^n$, so close that the two curves will wind around the orgin the same number of times. Intuitively, imagine you are out walking a dog on a leash. If you make sure that the leash is shorter than half the distance to a flagpole at each time, there is no risk that the dog will get tangled up.
