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.

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2 Responses to “The Argument Principle, The Winding Number, and Rouché’s Theorem (Part 2.)”

  1. Anonymous said

    you miss the proof

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