I'm trying to prove FTA by using the maximum principle. Here's what I did,

Let $P$ be a polynomial of degree at least $1$ and assume that $P$ has no zeros. Define $$f(z):=\frac{1}{P(z)}.$$ Then $f$ is holomorphic on the disk $|z| \leq R$. Since $f$ is continuous, it attains its maximum value for some complex number, say $w$. By the Maximum Principle, $w$ lies on the boundary and $f(z) \leq f(w)$ for all $|z| \leq R$

How do I get a contradiction from here?

Thank you


1 Answer 1


You can't say $f(z) \le f(w)$, because these are complex numbers; what you want is $|f(z)| \le |f(w)|$.

Now use the fact that $|P(z)| \to \infty$ as $|z| \to \infty$.

  • $\begingroup$ Sorry, I forgot to put in the absolute value. $\endgroup$
    – Sai
    Commented Jan 26, 2015 at 7:47
  • $\begingroup$ So if |p(z)| goes to infinity then |f(z)| goes to zero. Which shows that |p(w)| is greater than or equal to zero. $\endgroup$
    – Sai
    Commented Jan 26, 2015 at 7:50
  • 2
    $\begingroup$ The point is that $|f(w)|$ goes to $0$ as $R \to \infty$. What does that say about $f(z)$? $\endgroup$ Commented Jan 26, 2015 at 16:07
  • $\begingroup$ Sorry but why $|f(w)|$ goes to zero? And what it says about $f(z)$? $\endgroup$
    – RFZ
    Commented Apr 17, 2019 at 16:33
  • $\begingroup$ If $w_R$ is a point in the disk $|z|\le R$ that maximizes $|f|$ there, and $R < S$, then $|f(w_R)| \le |f(w_S)|$ (because the maximization over $|z| \le S$ includes $w_R$). But the Maximum Principle says $|w_R| = R$, and $|f(w_R)| = \max_{|z|=R} |f(z)| \to 0$ as $R \to \infty$. You can't have a positive, increasing function that goes to $0$. $\endgroup$ Commented Apr 17, 2019 at 17:52

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