## The Argument Principle, The Winding Number, and Rouché’s Theorem (Part 2.)

### December 15, 2010

(I’ve decided against giving a proof of Rouché’s theorem until such a time as I find one that doesn’t use algebraic topology or isn’t tedious as hell.)

Let’s simply state Rouché’s theorem, and then we’ll talk about how to actually apply Rouché’s theorem.

Theorem (Rouché): If $f, g$ are meromorphic functions (holomorphic except at a finite number of finite poles) inside a simple closed curve $C$ which have no zeros or poles on $C$.  If we have that

$|f(z) + g(z)| < |f(z)| + |g(z)|$

holds for all $z$ on the boundary of $C$ then:

$Z_{f} - P_{f} = Z_{g} - P_{g}$

where $Z_{f}$ denotes the number of zeros of $f$ counted with multiplicity in the interior of $C$ and $P_{f}$ denotes the number of poles of $f$ counted with orders in the interior of $C$

In other words, if those two assumptions hold, the difference of their zeros and poles are the same on the interior of $C$.

A direct and useful corollary of this theorem is what happens if the meromorphic function has no poles; ie, if our function is actually holomorphic:

Corollary: With the assumption above, if $f$ and $g$ are entire (ie, have no poles and are holomorphic) and we have that

$|f(z) + g(z)| < |f(z)| + |g(z)|$

on the boundary of $C$, then the two functions share the same number of zeros with multiplicity.

## That’s cool and all, but how can I use it?

Let’s do a few examples to really show you the power of this theorem!

Example 1.  Let’s start with an easy one.  How many zeros does $z^{2} + 0.5z$ have in the unit disk?

Solution 1.  If you didn’t know the theorem above, what could we do?  Factor, of course!  $f(z) = z(z+0.5)$ and so it has zeros at $z = 0$ and $z = 0.5$.  Thus, it has two zeros in the unit disk.

Alternatively, if you want to apply the theorem, we let $g(z) = -z^{2}$.  See the note in Solution 2 as to why we picked this: we essentially look for the term with the greatest coefficient, because it makes calculations a lot easier.  Then,

$|f(z) + g(z)| = |z^{2} + 0.5z - z^{2}| = |0.5z| = 0.5|z| = 0.5$

on the unit disk’s boundary, the unit circle where $|z| = 1$.  Now note,

$|f(z)| + |g(z)| > |g(z)| = |-z^{2}| = |z^{2}| = |z|^{2} = 1$

and as $1 > 0.5$ this implies $|f(z) + g(z)| < |f(z)| + |g(z)|$ which implies that $f$ has the same number of zeros as $g$.  Since $g$ has two zeros (with multiplicity!) on the unit disk (namely, $z = 0$ with multiplicity 2) so does $f$.  Notice here that the zeros of $g$ don’t really tell us anything specific about where the zeros are; they just tell us about how many there are within a certain bound.  This idea is important in the previous post’s theorems, and, you know, sometimes it’s nice to know how many zeros we’re working with.

Example 2.  Given the polynomial $f(z) = z^{8} + 6z^{4} - z - 1$, how many zeros does this function have inside the unit disc?

Solution 2.  If you didn’t know the theorem above, this would be a rather difficult problem!  It takes a bit of fiddling around to figure out what function we want to use, but it usually is going to be one or a few terms of our original function negated.  We also usually use the ones with big coefficients.  Doing a few problems like this will lead you to why such a criteria is useful.  Either way, let’s let $g(z) = -6z^{4}$

$|f(z) + g(z)| = |z^{8} + 6z^{4} - z - 1 - 6z^{4}|$

$= |z^{8} - z - 1| \leq |z^{8}| + |z| + |1| = |z|^{8} + |z| + 1 = 3$

as $z$ is on the boundary of our unit disc (the unit circle) and so $|z| = 1$.  Now, note that

$|f(z)| + |g(z)| = |z^{8} + 6z^{4} - z - 1| + |-6z^{4}|$

$> |-6z^{4}| = 6|z|^{4} = 6$

and since $6 > 3$, it follows that $|f(z) + g(z)| < |f(z)| + |g(z)|$.  It follows that $f$ has the same number of zeros in the unit disk as $g$ does; by the fundamental theorem of calculus and some easy factoring, we find that $g$ has four zeros in the unit disk.  Indeed, this is true.

Example 3.  How many zeros does $f(z) = z^{5} - 6z^{4} + z^{3} + z^{2} + z + 1$ have in the unit disk?

Solution 3.  This function is crazy!  But, let’s use the theorem and choose some function using the biggest coefficients of $f$.  In this case, let’s let $g(z) = 6z^{4}$.  Then,

$|f(z) + g(z)| = |z^{5} - 6z^{4} + z^{3} + z^{2} + z + 1 + 6z^{4}|$

$= |z^{5} + z^{3} + z^{2} + z + 1|$

$\leq |z|^{5} + |z|^{3} + |z|^{2} + |z| + 1 = 1 + 1 + 1 + 1 + 1 = 5$

$|f(z)| + |g(z)| < |g(z)| = 6|z|^{4} = 6$

and since $6 > 5$ we have that $|f(z) + g(z)| < |f(z)| + |g(z)|$.  Applying the theorem, we find that $f$ has exactly four zeros in the unit disk.  Indeed, this is true!

Notice something here: we’ve used a lot that $|f(z)| + |g(z)| < |g(z)|$ which gives us an easy way to bound the second inequality; in fact, this is true if $|f(z)|$ is non-zero on the boundary of the curve!  Neat.