In the text "Function Theory of One Complex Varible", by Robert E.Greene and Steven G.Krantz i'm inquiring if I completed the proof sketch for $\text{Theorem (1)}$ correctly and can one provide a hint for a simpler proof ? The initial sketch by the authors is given in the section called $\text{Sketch}$ and my initial solution is given in the section titled $\text{Proof}$

$\text{Theorem 1} \, \, (\text{Fundamental Theorem of Algebra}).$ Given any positive integer $n \geq 1$ and choice of complex numbers $a_{o}, a_{1},...,a_{n}$, such that $a_{n} \neq 0$, the polynomial equation

$$a_{n}z^{n} + \cdot \cdot \cdot + a_{1}z + a_{o} = 0$$

has at least one solution $z \in \mathbb{C}$


If $p(z)$ is a nonconstant polynomial and $p(z)$ never vanishes, then $H(z) \equiv 1/p(z)$ is an entire function( hence, in particular, it is continuous) which vanishes at $\infty$. Therefore $H$ has a maximum at some point $P_{0} \in \mathbb{C}$.Examine the Taylor expansion of $H$ about $P_{0}$ to see that this conclusion is impossible


Considering our Sketch of our trivial proof to $\text{Theorem (1)}$, in order to reach practical results we will need to fully to define $H(z) \equiv 1/p(z)$ and construct it's trivial Taylor Expansion the proceeding developments can be followed in $\text{Lemma (1)}.$

$\text{Lemma (1)}$

Let $U \subset \mathbb{C}$ be an open set such that $H: D(P_{0},r) \rightarrow \mathbb{C}$ note that $D(P_{0},r) \subset \mathbb{C}$. Utilizing $\text{Theorem (1.2)}$ we arrive at the trivial construction of a Taylor expansion for $H(z)$.

$\text{Theorem (1.2) }$

Let $U \subset \mathbb{C}$ be an open set and let f be holomorphic on $U$. Let $P \in U$ and suppose that $D(P,r) \subset U$. Then the Taylor Expansion has a radius of convergence of at least $r$. It converges to $f(z)$ on $D(P,r)$.

$$f(z) = \sum_{k=0}^{\infty}\frac{((\partial_{z}f)^{k}) (P_{0})}{k!}(z-P)^{k}$$


$$ H(z) = \sum_{K} \bigg( \partial_{z}^{k} H(P_{0}) \bigg) \frac{(z-P_{0})^{K}}{K!} $$

$\text{Lemma (2)}$

Our Taylor Expansion for $H(z)$ is fully achieved in $(2)$, using this what we've achieved, we arrive at the formulation of the claim


$$H(z) = \sum_{K} \bigg( D_{z}^{k} \frac{1}{a_{n}P_{0}^{n} \bigg( \frac{a_{0}}{a_{n}P_{0}^{n}} + \cdot \cdot \cdot + 1 \bigg)} \bigg) (z-P_{0})^{K}/{K!}.$$


One must formally show for $a_{n}=0$ for $n≤1$ by estimating

$$|a_n| = \, \left| \frac{1}{2 \pi i} \oint_{\partial D(P_{0},r) } \frac{H(\zeta)}{(\zeta - P_{0})^{n+1} } \, d \zeta \right| \leq \frac{1}{2 \pi} \frac{M}{R^{n+1}}2 \pi R.$$

$\text{Lemma (3)}$

To verify the recently conjectured claim, one must rely on the Cauchy Estimates as formally discussed utilizing the result one reaches the following developments

$\text{Theorem (1.3) The Cauchy Estimates}$

Let $f:U \rightarrow \mathbb{C}$ be a holomorphic function on an open set $U$, $P \in U$, and assume that the closed disc $\overline D(P,r), r > 0$ is contained in $U$. Set $M = \sup_{z \in \overline D(P,r)} |f(z)|$. Then for $k = 1,2,3..$ we have

$$ \bigg| \partial_{z}^{k}f(P) \bigg | \leq \frac{\sup_{z \in \overline D(P,r)} |f(z)| k!}{r^{k}}$$


One can notice, we can get the following power series from $(2.2)$

$$H(z)= \sum_{K} \frac{1}{2 \pi i}\oint_{\partial D(P,r)} \frac{H(\zeta)}{(\zeta - P_{0})^{n+1} } \, d \zeta(z-P)^{n}.$$

Applying $(1.3)$ to our recent developments yields the following


$$|a_n|= \left| \frac{((\partial_{z}f)^{k}) (P_{0})}{k!} \right| = \left| \frac{1}{2 \pi i} \oint_{\partial D(P_{0},R) } \frac{H(\zeta)}{(\zeta - P_{0})^{n+1} } \, d \zeta \right| \leq \frac{n! \sup_{z \in \overline{D(P,r)}} |H(z)|}{r^n}.$$

Utilizing the triangle inequality one can arrive at the final conclusion in $(6.6)$


$$\left| \frac{1}{2 \pi i} \oint_{\partial D(P_{0},r) } \frac{H(\zeta)}{(\zeta - P_{0})^{n+1} } \, d \zeta \right| \leq \left| \frac{1}{2 \pi i} \right| \oint_{\partial D (P_{0},r)} \left| \frac{H(\zeta)}{(\zeta - P_{0})^{n+1}} \right| \left| d \zeta \right|$$

$$ \, \, \, \, \, \, \, \, = \frac{1}{2 \pi i} \oint_{\partial D(P_{0},r)} \frac{|H(\zeta)|}{\left| (\zeta - P_{0})^{n+1} \right|} \left| d \zeta \right| = \frac{1}{2 \pi} \frac{M}{R^{n+1}}2 \pi R$$

In summary, since $|a_{n}| \leq KR^{-n}$ for every $R$ each $n >0$ by letting $R \rightarrow \infty \, \, H'(z) = 0$.

  • $\begingroup$ Your statement of the FTA is weird. Positive integers are $\ge 1$ and why do you say $a_n\le 0?$ $\endgroup$
    – zhw.
    Jun 24, 2018 at 16:28
  • $\begingroup$ @zhw ahh okay thanks for finding the typo i'll have to fix that $\endgroup$
    – Zophikel
    Jun 24, 2018 at 16:30
  • 2
    $\begingroup$ What's the maximum of a complex function to begin with? $\endgroup$
    – egreg
    Jun 24, 2018 at 16:36

1 Answer 1


Your proof may well be correct, but I did not read it carefully. I'm pretty sure, however, that the authors did not intend for you to essentially prove Liouville's theorem along the way.

Here's a simpler approach, which is a local argument: We don't even need the function $H;$ let's go back to the polynomial $p,$ which is nonconstant and such that $|p|$ has a positive minimum at $z_0.$ We can write $p(z) = p(z_0) +a_m(z-z_0)^m + \cdots + a_{n}(z-z_0)^{n} ,$ where $a_m\ne 0.$ This gives

$$\tag 1 p(z_0+re^{it}) = p(z_0) + a_mr^me^{imt}+ O(r^{m+1}) $$

as $r\to 0^+.$ Now show this leads to a contradiction by choosing a direction $t$ such that for small $r,$ $(1)$ has modulus $<|p(z_0)|.$

  • $\begingroup$ That may be true the proof is correct and what I did is unnecessary, what I initially did was take the same approach to prove Liouville's theorem and applied it here, also it seems one can use the Maximum Modules Principle to arrive at a given contradiction. $\endgroup$
    – Zophikel
    Jun 27, 2018 at 19:58
  • $\begingroup$ If you have covered Liouville, then why not just apply it to $H$? $\endgroup$
    – zhw.
    Jun 27, 2018 at 20:03
  • $\begingroup$ I'm on moblie atm at first when looking at your solution I thought your exploiting Liouville's Theorem in some form but now I realize this is not the case, hence why I asked about the Maximum modules principle. $\endgroup$
    – Zophikel
    Jun 27, 2018 at 20:07
  • $\begingroup$ No, my proof is elementary and really more of a real analysis proof. It does not depend on any of the magic of complex analysis. But yes, if you can use either Liouville or MMP, a very quick proof can be found. $\endgroup$
    – zhw.
    Jun 27, 2018 at 20:27
  • $\begingroup$ Thanks @zhw i'll be able to get 1 or 2 answers sometime this week :), thanks for begin patient with me $\endgroup$
    – Zophikel
    Jun 27, 2018 at 20:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .