## 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 are meromorphic functions (holomorphic except at a finite number of finite poles) inside a simple closed curve which have no zeros or poles on . If we have that

holds for all **on the boundary of ** then:

where denotes the number of zeros of counted with multiplicity in the interior of and denotes the number of poles of counted with orders in the interior of .

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

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 and are entire (ie, have no poles and are holomorphic) and we have that

on the boundary of , 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 have in the unit disk?

**Solution 1. **If you didn’t know the theorem above, what could we do? Factor, of course! and so it has zeros at and . Thus, it has two zeros in the unit disk.

Alternatively, if you want to apply the theorem, we let . 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,

on the unit disk’s boundary, the unit circle where . Now note,

and as this implies which implies that has the same number of zeros as . Since has two zeros (with multiplicity!) on the unit disk (namely, with multiplicity 2) so does . Notice here that the zeros of 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 , 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 .

as is on the boundary of our unit disc (the unit circle) and so . Now, note that

and since , it follows that . It follows that has the same number of zeros in the unit disk as does; by the fundamental theorem of calculus and some easy factoring, we find that has four zeros in the unit disk. Indeed, this is true.

**Example 3.** How many zeros does 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 . In this case, let’s let . Then,

and since we have that . Applying the theorem, we find that has exactly four zeros in the unit disk. Indeed, this is true!

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

nice

you miss the proof