Geometric Complex Analysis: Rouché’s Theorem

Complex analysis is one of the most rich and amazing areas of mathematics, for the complex numbers $\mathbb{C}$ possess geometric properties that intertwine with their analytic properties in surprising and beautiful ways. This is hardly a place to discuss the subject in detail, so I will focus on Rouché’s Theorem, with an entertaining application.

Rouché’s Theorem If $U$ is an open disc in $\mathbb{C}$ , and $f$ and $g$ are complex-valued differentiable functions defined in some neighborhood of $\overline{U}$ , and if $|f+g| < |g|$ on the boundary of $U$ , then $f$ and $g$ have the same number of zeros inside $U$ , up to multiplicity.

By “multiplicity” here we mean the following: $\alpha$ is a zero of multiplicity 1 of $h$ if $h(\alpha)=0$ but $h'(\alpha)\neq 0$ , and a zero of multiplicity two if $h(\alpha)=h'(\alpha)=0$ but $h''(\alpha)\neq 0$ , etc.

The Wikipedia article on this theorem has an interesting informal summary: “If a person were to walk a dog on a leash around and around a tree, and the length of the leash is less than the radius of the tree, then the person and the dog go around the tree an equal number of times.” A little bit of an oversimplification perhaps, but an alluring hint at the deep geometric principles at work in this theorem.

An application:

Proposition: For any $n=2,3,\ldots$ , the polynomial $p_n(x) = 1 - x - x^2 - \ldots - x^n$ is irreducible.

Proof: Let $f(x) = (1-x)p_n(x) = x^{n+1} - 2x + 1$ , and let $g(x) = 2x$ . Choose $r < 1$ to be sufficiently close to $1$ that $r^{n+1}+1 < 2r$ .

Then, when $|x| = r$ , we have $|f(x) + g(x)| = |x^{n+1} + 1| \leq r^{n+1} + 1 \leq 2r = |2x| = |g(x)|$ , so $f$ and $g$ satisfy the conditions of Rouché’s Theorem, and have the same number of zeros inside the circle around the origin of radius $r$ . Letting $r\to 1^-$ , we find that $f$ and $g$ have the same number of roots inside the circle of radius $1$ .

But $g$ has exactly one root inside this circle, so $f$ , hence $p_n(x)$ must have exactly one root $\alpha$ with $|\alpha| < 1$ .

Now, suppose that $\beta$ were a root of $f(x)$ with $|\beta| = 1$ . Then $\beta - 1/2 = \beta^{n+1}/2$ , so $|\beta - 1/2| = 1/2$ . So $\beta$ lies on a circle around $1/2$ of radius $1/2$ , as well as on the circle around the origin of radius $1$ , so $\beta=1$ . Since $p_n(1) = 1-n\neq 0$ , $p_n(x)$ has no roots on the unit circle.

Now we are done! For if $p_n(x) = a(x)b(x)$ over the integers, then the constant terms of $a$ and $b$ must be $\pm 1$ . But these constant terms are the product of some subset of the roots of $p_n$ . If $\alpha$ is a root of $a$ , since it is the only root of absolute value less than or equal to one, it follows that $a$ must have every single other root of $p_n$ as a root, and the factorization must be trivial.

For anyone interested in studying this subject in greater detail, I highly recommend Functions of One Complex Variable by John B. Conway. And if you want to know more about numbers like our $\alpha$ above, read up on Pisot numbers.

This entry was posted on Friday, October 3rd, 2008 at 6:42 am and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Geometric Complex Analysis: Rouché’s Theorem”

antimatroid Says:
October 3, 2008 at 8:09 pm | Reply
Another recommendable text on the subject is
Complex Variables: Introduction and Applications by Ablowitz and Fokas.

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Geometric Complex Analysis: Rouché’s Theorem

One Response to “Geometric Complex Analysis: Rouché’s Theorem”

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