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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 ?

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You don't know that $P(z)$ can be factored at all. What if $P(z) = g(z)$? I would instead start by supposing that $P(z)$ has no zeros, and reach a contradiction by showing that integral is nonzero. If $P(z)$ has at least one zero, then it has $n$ zeros by simply iterating. –  Christopher A. Wong Sep 19 '12 at 7:03
    
I would also add, now that I'm looking at your argument more closely - when you look at the term $g'(z)/g(z)$, it is true that the term becomes small when $R$ becomes large, but then remember that you are simultaneously integrating around a larger and larger circle, so the reasoning you gave doesn't work. –  Christopher A. Wong Sep 19 '12 at 7:21
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2 Answers 2

up vote 4 down vote accepted

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.

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thanks. I don't quite see it yet. I understand that as $|z|=R\to \infty$, the dominant term is $a_n z^n$, which rotates $z$ around the origin $n$ times. How do I formalize it ? –  Teddy Sep 20 '12 at 7:45
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You can formalize it by performing an estimate on the difference between the integral of $f'(z)/f(z)$ and the integral of $g'(z)/g(z)$ where $g(z) = a_n z^n$ is the dominant term. Using the estimate you can show that difference in integrals is small, but you know that each integral can only take a multiple of $2\pi i$ as its value according to the argument principle, so they must be equal. –  Christopher A. Wong Sep 20 '12 at 20:55
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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 $(n+1)$.

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