The Argument Principle used to prove the Fundamental Theorem of Algebra Greene and Krantz pose the following problem in Function Theory of One Complex Variable, Ch. 5 problem 3:

Give another proof of the fundamental theorem of algebra as follows:
  Let $P(z)$ be a non-constant polynomial. Fix $Q\in \mathbb{C}$.
  Consider 
  \begin{equation}
 \frac{1}{2\pi i} \oint_{\partial D(Q,R)} \frac{P'(z)}{P(z)}\,dz. 
\end{equation}
  Argue that as $R\to +\infty$, this
  expression tends to a nonzero constant.

I was thinking along these lines: Since we do not know $P(z)$ factors completely, let us write
$$ P(z) = \prod_j (z - \alpha_j) \, g(z),$$
where $g(z)$ is an irreducible polynomial. Now
$$ \frac{P'(z)}{P(z)} = \sum_k \frac{1}{z-\alpha_k} + \frac{g'(z)}{g(z)}.$$
Each of the terms $1/(z-\alpha_k)$ adds $1$ to the integral expression. As $R \to \infty$, all the $\alpha_k$ are eventually inside $D(Q,R)$, whereas the term $g'(z)/g(z)$ approaches zero, since the denominator has a higher degree.
Is the reasoning correct ? Can someone offer a simpler argument ?
 A: Here is a general hint that should lead to a short and nice proof: the argument principle tells you that the integral
$$ \frac{1}{2\pi i} \int_{\Gamma} \frac{f'(z)}{f(z)} \, dz = \#(zeros) - \#(poles) $$
but what is more important is what the above integral represents. Note that $f'(z)/f(z) = (\log(f(z))'$, the above integral (without the $2\pi i$ factor) tells you the total change in the complex argument (i.e. angle) of the values of $f(z)$ as you traverse the contour $\Gamma$. Now, for $|z| = R$ very large, think about what happens to the angles of a polynomial $P(z) = a_n z^n + \ldots + a_1 z + a_0$; this should allow you to compute the above integral directly.
A: Let me sum things up. Let $$ P(z) = \sum_{j=1}^n a_j z^j $$
be of $n$-th degree. We do know that a polynomial of degree $n$ has at most $n$ roots.
Since the number of roots is finite, we may choose $R$ such that $D(Q,R)$ contains all the roots, and $R>|Q|$. (we do not yet know there are roots. This is just what the theorem says, in fact - that there are roots.)
For $r>R$, let us look at the integral 
\begin{equation}
 \frac{1}{2\pi i} \oint_{\partial D(Q,r)} \frac{P'(z)}{P(z)}\,dz. 
\end{equation}
On the one hand, this equals the number of zeros of $P(z)$ inside $D(Q,r)$. Since $r>R$, this is the total number of zeros, which we'll denote $N$.
On the other hand, let us calculate that same integral for a simpler polynomial, a monomial in fact, $g(z) = a_n z^n$, the leading term in $P(z)$.
\begin{equation}
 \frac{1}{2\pi i} \oint_{\partial D(Q,r)} \frac{g'(z)}{g(z)}\,dz = \frac{1}{2\pi i} \oint_{\partial D(Q,r)} \frac{n a_n z^{n-1}}{a_n z^n}\,dz = \frac{1}{2\pi i} \oint_{\partial D(Q,r)} \frac{n}{z}\,dz =n,
\end{equation}
where the last equality follows from (e.g.) the residue theorem (here the assumption $R>|Q|$ was used, to get $0$ inside the integration path).
Finally, we'd like to show that the number of zeros of $P(z)$, which we denoted $N$, is equal to the degree of $P$, namely $n$. To this end, we'll show 
$$  \underbrace{\frac{1}{2\pi i} \oint_{\partial D(Q,r)} \frac{P'(z)}{P(z)}\,dz}_\textrm{N} - \underbrace{\frac{1}{2\pi i} \oint_{\partial D(Q,r)} \frac{n}{z}\,dz}_\textrm{n} =0.$$
Here's the thing:
\begin{equation}
\frac{1}{2\pi i} \oint_{\partial D(Q,r)} \frac{P'(z)}{P(z)}- \frac{n}{z} \,dz = 
\frac{1}{2\pi i} \oint_{\partial D(Q,r)} \frac{z P'(z) - n P(z)}{z P(z)} \,dz \xrightarrow[r\to \infty]{} 0
\end{equation}
where the limit follows from the fact that the numerator is an $(n-1)$ degree polynomial, and the denominator is of degree $(n+1)$.
