# Proving the Fundamental Theorem of Algebra via Taylor Series?

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 $$z \in \mathbb{C}$$

$$\text{Sketch}$$

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

$$\text{Proof}$$

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 $$D(P,r)$$.

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

$$(1.1)$$

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

$$\text{Lemma (2)}$$

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

$$(2.2)$$

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

$$(3.3)$$

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

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

$$(4.4)$$

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

$$(5.5)$$

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

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

$$(6.6)$$

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

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

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

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

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 $<|p(z_0)|.$

• 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. Jun 27, 2018 at 19:58
• If you have covered Liouville, then why not just apply it to $H$?
– zhw.
Jun 27, 2018 at 20:03
• 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. Jun 27, 2018 at 20:07
• 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.
– zhw.
Jun 27, 2018 at 20:27
• Thanks @zhw i'll be able to get 1 or 2 answers sometime this week :), thanks for begin patient with me Jun 27, 2018 at 20:44