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 ?
Best Answer
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.