The Fundamental Theorem of Algebra

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We investigate Gauss’ first proof of the Fundamental Theorem of Algebra.

Gauss’ First Proof

In 1799, as his doctoral dissertation, Gauss gave this proof of the Fundamental Theorem of Algebra (FTA).

Remind me: what does the FTA say?

To work through this proof, we will need DeMoivre’s Theorem, which says that $(\cos (\theta ) + i \sin (\theta ))^n = \cos (n\theta ) + i \sin (n\theta ).$

Explain why DeMoivre’s Theorem tells us how to find the $n$ -th roots of any complex number.

First, Gauss writes his general polynomial (with real coefficients) as $X = x^m + A x^{m-1} + B x^{m-2} + \dots + L x + M.$ He is trying to show that $X$ has either a real factor $x-r$ or a complex factor $x^2 - 2xr\cos (\phi ) + r^2$ .

Why is the above statement the same as saying $X$ has $x=r(\cos (\phi ) + i\sin (\phi ))$ as a root?

Now, Gauss plugs $x=r(\cos (\phi ) + i\sin (\phi ))$ in to $X$ and separates his result into its real part and its imaginary part, getting $U = r^m\cos (m\phi ) + A r^{m-1} \cos ((m-1)\phi ) + \dots + L r \cos (\phi ) + M$ and $T = r^m\sin (m\phi ) + A r^{m-1} \sin ((m-1)\phi ) + \dots + L r \sin (\phi ).$

How was DeMoivre’s Theorem involved, here? Which of $U$ and $T$ is the real part? The imaginary part?

Gauss then proves directly that if $T=U=0$ , then either $X$ is divisible by $x-r$ or $X$ is divisible by $x^2-2xr\cos (\phi ) + r^2$ . We won’t prove that here, but keep in mind that means we are trying to prove that the graphs of $T=0$ and $U=0$ have an intersection.

Plot $T=0$ , $U=0$ and a very large circle of radius $R$ on the polar axes (i.e. $r$ and $\phi$ ). As $R\to \infty$ , the curves $T=0$ and $U=0$ look more and more like $\text {Re}(x^m)$ and $\text {Im}(x^m)$ .

Why is this last statement true? What does it have to do with DeMoivre’s Theorem?

The upshot of the last point is: if we draw a circle with large enough radius $R$ , and look at the pieces (more technically “branches”) of $T=0$ and $U=0$ behave inside our circle, we find that the intersections of the branches with the circle itself must alternate.

Why does the horizontal axis ( $\phi = 0$ and $\phi = \pi$ ) have to be a branch of $T$ ?

On the circle below, sketch the horizontal axis and label it $T=0$ . Since each branch must both enter and leave the circle, label an even number of $U$ -intersections, alternating with an even number of $T$ -intersections. Be sure to label which is which!

Why are we now forced to have an intersection of one of the branches of $T$ with one of the branches of $U$ ? (It may help to use two different colors!)

That’s it!

d’Alembert’s Proof

Here is a theorem now known as d’Alembert’s Lemma:

If $p(z)$ is a polynomial function with $p(z_0) \neq 0$ for some $z_0$ , then any neighborhood of $z_0$ contains a point $z_1$ such that $\vert p(z_1) \vert < \vert p(z_0) \vert$ .

What is this lemma saying?

We’ll also need the Extreme Value Theorem from calculus (sometimes called the Weierstrass Extreme Value Theorem).

Remind me: what does the Extreme Value Theorem say about a continuous function on a closed interval?

On to the proof of the FTA. Step 1: We can find some radius $R$ so that for all $\vert z \vert \geq R$ , $\vert p(z) \vert$ is an increasing function.

What is Step 1 saying? Why is it true? (Hint: what does the graph of a degree- $m$ polynomial look like? What about the absolute value of that polynomial?

Step 2: We know that $\vert p(z) \vert$ has to have an absolute minimum inside the circle.

What is Step 2 saying? Why is it true?

Step 3: Suppose the minimum is strictly greater than zero.

If the minimum is strictly inside the circle, why does d’Alembert’s lemma give us a contradiction?

If the minimum is on the circle’s boundary, why do we have a contradiction?

Why have we now proven the FTA?