# Ngô Quốc Anh

## May 17, 2009

### Rouché’s theorem and several applications

Filed under: Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 16:51

In mathematics, especially complex analysis, Rouché’s theorem tells us that if the complex-valued functions $f$ and $g$ are holomorphic inside and on some closed contour $C$, with $|g(z)|<|f(z)|$ on $C$, then $f$ and $f + g$ have the same number of zeros inside $C$, where each zero is counted as many times as its multiplicity. This theorem assumes that the contour $C$ is simple, that is, without self-intersections. Moreover, $f$ and $f + g$ should have no zeros on $C$.

The theorem is usually used to simplify the problem of locating zeros, as follows. Given an analytic function, we write it as the sum of two parts, one of which is simpler and grows faster than (thus dominates) the other part. We can then locate the zeros by looking at only the dominating part.

For example, the polynomial $z^5+3z^3+7$ has exactly $5$ zeros in the disk $|z|<2$ since $|3z^3+7|<32 = | z^5 |$ for every $| z | = 2$, and $z^5$, the dominating part, has five zeros in the disk.

Geometric explanation

It is possible to provide an informal explanation on why the Rouche’s theorem holds. First we need to rephrase the theorem a little bit. Let $h(z) = f(z) + g(z)$. Notice that $f, g$ holomorphic implies $h$ holomorphic too. Then, with the conditions imposed above, Rouche’s theorem says that

If $|f(z)| > |h(z)-f(z)|$ then $f(z)$ and $h(z)$ have the same number of zeros on the interior of $C$.

Notice that the condition $|f(z)| > |h(z)-f(z)|$ means that for any $z$, the distance of $f(z)$ to the origin is larger than the length of $h(z)- f(z)$, which in the following picture means that for each point on the blue curve, the segment joining to the origin is larger than the green segment associated to it. Informally we can say that the red curve $h(z)$ is always closer to the blue curve $f(z)$ than to the origin.

But the previous paragraph shows that since $f(z)$ winds exactly once around $0$, so must $h(z)$, and by the argument principle, the index of both curves around zero is the same, which means that $f(z)$ and $h(z)$ have the same number of zeros.

One popular, informal way to summarize this argument is as follows: 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. (Indeed, one may see that the converse of Rouche’s theorem is false, insofar as the leash need only be less than the circumference of the tree.)

Applications

Consider the polynomial $z^2 + 2az + b^2$ (where $a > b > 0$). By the quadratic formula we find that it has two zeros at $a \pm \sqrt{a^2 - b^2}$. Since

$\displaystyle |z^2 + b^2| \le 2b^2 < 2a|z|$

for every $| z | = b$. Rouché’s theorem says that the polynomial has exactly one zero inside the disk $| z | < b$. Since $a + \sqrt{a^2 - b^2}$ is clearly outside the disk, we conclude that the polynomial has a zero at $a - \sqrt{a^2 - b^2}$. This sort of arguments can be useful in locating residues when one applies Cauchy’s Residue theorem.

Rouché’s theorem can also be used to give a short proof of the Fundamental Theorem of Algebra. Let $p(z) = a_0 + a_1z + a_2z^2 + ... + a_nz^n$, and choose a $R$ so large that

$\displaystyle |a_0 + a_1z + ... + a_{n-1}z^{n-1}| \le \sum_{j=1}^n |a_j| R^{n-1} < |a_n|R^n = |a_n z^n|$

for every $| z | = R$.  Since $a_nz^n$ has $n$ zeros inside the disk $| z | < R$, it follows from Rouché’s theorem that $p$ also has the same number of zeros inside the disk.

One advantage of this proof over the others is that it shows not only that a polynomial must have a zero but the number of its zeros is equal to its degree (counting, as usual, multiplicity).

Another use of Rouché’s theorem is to prove the open mapping theorem for analytic functions.

Proof of Rouché’s theorem

The hypothesis, that $|g(z)| < |f(z)|$ on $C$, implies

$\displaystyle\left|\frac{f(z)+g(z)}{f(z)}-1\right| < 1$

for all $z \in C$. Hence the function $F(z) = \frac{f(z)+g(z)}{f(z)}$ takes the curve $C$ to a curve $F(C)$ in the interior of the disc of radius $1$ and center $1$. The winding number of $F(C)$ about the origin is thus zero. On the other hand, by the argument principle, this winding number is given by

$\displaystyle 0 = {1\over 2\pi i}\oint_C {F'(z) \over F(z)} dz=N_F(C)-P_F(C)$

where $N_F(C)$ is the number of zeroes of $F$ inside $C$, $P_F(C)$ is the number of poles inside $C$. Hence $N_F = P_F$. But $F$ is the ratio of two holomorphic functions $f+g$ and $f$ inside $C$, and so the zeros are those of $f+g$ and the poles are the zeros of $f$. That is,

$\displaystyle 0=N_F(C) - P_F(C) = N_{f+g}(C) - N_f(C)$,

as required.

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