The Argument Principle used to prove the Fundamental Theorem of Algebra

Question

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 ?

Answer 1

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 $(n+1)$.

Answer 2

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.

Answer 3

Old question but here's another simple argument .

Well , Rouche's Theorem is a consequence of the Argument Principle. If $f(z)=a_{n}z^{n}$ and $g(z)=\sum_{k=0}^{n-1}a_{k}z^{k}$ where $|a_{n}|\neq 0$ then for a circle $C_{R}$ of large enough radius $R$ , we have that $|f(z)|>|g(z)|$ for all $z\in C_{R}$.

So $f(z)$ and $f(z)+g(z)=p(z)$ have the same number of zeroes.

But again by argument principle the number of zeroes of $f$ inside $C_R$ is given by $\displaystyle\frac{1}{2i\pi}\oint_{C_{R}}\frac{f'(z)}{f(z)}\,dz = \frac{1}{2i\pi}\oint\frac{n}{z}\,dz = n$ . And the number of zeroes remains constant$(=n)$ for all large $R$. So , the number of zeroes of $f$ in the complex plane is $n$ and hence $p(z)$ also has $n$ zeroes.

(The above is just a flex really. Obviously $a_{n}z^{n}$ has $0$ as it's root with multiplicity $n$).

Answer 4

How about this

Suppose $f(z)=z^n+a_{n-1}z^{n-1}+..+a_1z+a_0$. For $0\leq t \leq 1$ denote; $ F_t=z^n+t(a_{n-1}z^{n-1}+..+a_1z+a_0) $ Then for any circle $C_r$ of radius $r>1$ we have \begin{align*} |F_t(z)|\geq |z|^n-|t||a_{n-1}z^{n-1}+..+a_1z+a_0| \\ \geq 1 - (|a_{n-1}|+..+|a_1|+|a_0|) \end{align*}

Notice that we may assume that $|a_{n-1}|+..|a_1|+|a_0|<1$ as otherwise we may consider $z=cw$ in $z^n+a_{n-1}z^{n-1}+..+a_1z+a_0=0$ which gives us, $ w^n+\frac{a_{n-1}}{c}z^{n-1}+..+\frac{a_1}{c^{n-1}}+\frac{a_0}{c^n}=0 $ Choose $c$ with $|c|$ large enough so that $|\frac{a_{n-1}}{c}|+..+|\frac{a_1}{c^{n-1}}|+|\frac{a_0}{c^n}|<1$. \par So we have $|F_t(z)|>0$. Now consider; $ n_t =\frac{1}{2\pi\sqrt{-1}} \int_{C_r} \frac{F'_t}{F_t}dz $ Now as $F_t$ is nowhere zero and holomorphic we have by argument principle that $n_t \in \mathbb{Z}$ (in particular $n_t$ denotes the number of zeros of $F_t$ in the interior of $C_r$). Now as $n_t$ is continuous we get $n_t$ is constant and in particular $n_0=n_1$. Now $n_1$ is the number of zeros of our $f(z)$ which is same as number of $z^n$ inside the interior of $C_r$. Hence $f$ has all its $n$ zeros inside the interior of $C_r$.