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Cool proof of the Fundamental Theorem of Algebra

Posted by on Jul 10, 2012 in Complex Analysis, Mathematics education | 0 comments

Despite being on holiday I can’t resist looking for cool proofs. This one is not so much cool as interesting in a Why-didn’t-I-think-of-that way. The fundamental theorem of algebra — that any polynomial has a complex root — is well known to be a theorem of analysis rather than algebra and many proofs are known. The proof I use in my course relies on Liouville’s Theorem and the Maximum Modulus Theorem. However, there is a more direct route using only Cauchy’s Integral Formula and the Estimation Lemma. I found the basic idea in Complex Proofs of Real Theorems by Peter Lax and Lawrence Zalcman. Here’s my modified version:

Every polynomial p(z)=a_nz^n+a_{n-1}z^{n-1} + \dots a_1z+a_0 , n\geq 1, a_n\neq 0, has a root in \mathbb{C} .

The proof is as follows.

Suppose not and derive a contradiction. Since p(z)\neq 0 for all z\in \mathbb{C} the function defined by 1/p(z) is differentiable on all of \mathbb{C} .
Now, for z\neq 0,


\left| \frac{p(z)}{z^n} \right| = \left| a_n+ \frac{a_{n-1}}{z} + \dots + \frac{a_0}{z^n} \right| \to |a_n| {\text{ as }} |z|\to \infty .

So there exists an R such that |z|\geq R implies that

\displaystyle \left| \frac{p(z)}{z^n} \right| \geq \frac{|a_n|}{2} .

Thus \displaystyle \left| \frac{1}{p(z)} \right| \leq \frac{2}{|a_n||z|^n} for such z.

As 1/p(z) is differentiable on all of \mathbb{C} we have for all r, by Cauchy’s Integral Formula, (and where C_r denotes a circle of radius r centred at the origin),


\begin{array}{rcl} \left| \dfrac{1}{p(0)} \right| &=& \left| \dfrac{1}{2\pi i } \displaystyle \int_{C_r} \dfrac{1/p(z)}{z-0} \, dz \right| \\ &\leq & \dfrac{1}{2\pi } \dfrac{2\times 2\pi r}{|a_n|r^n\times r } , {\text{ by the Estimation Lemma and for }} r>R . \end{array}

The right hand side can be made as small as we like by taking r large enough. Thus 1/p(0)=0. This is impossible so we have a contradiction and therefore p has a root.

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