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This past semester I took a graduate course in complex analysis which I completed moderately well in spite of my expectations (that is I honestly think I deserved a lower grade than I received). I had one assigned question which caused me to join MSE on the subject of Rouché's theorem. This problem comes from Chapter 5 section three of Conway's Functions of a single Complex Variable, the section is on the Argument Principle and Rouché's theorem. The problem is number 2 of this chapter and proceeds as such:

"Suppose $f$ is analytic on $\bar B(0 ; 1) $ and satisfies $\left| f(z) \right|<1 $ for $ |z|=1$. Find the number of solutions (counting multiplicities) of the equation $f(z) = z^n$ where $n$ is an integer larger than or equal to $1$."

Now my question isn't about solving this problem but more of a hint at where to start. As I understood Rouché's theorem, I am required to compare two functions, say $f$ and $g$, so I can compare the number of poles and zeroes. Every example I could find on MSE however included an equation with more than one term and seemed (at least to me) easier than the one I was given. I unfortunately do not have any work of my own to provide as I was definitely stumped. Any discussion of this will help, or references to MSE questions I might have missed would be greatly appreciated

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Perhaps this earlier question might help: math.stackexchange.com/questions/149752/… –  imranfat Jun 6 '13 at 21:49
    
Yeah I remember seeing a question similar to this, I actually have to admit that first thing I tried was defining a new function $g$ given by $ g = f - z^n$, but I didn't think this was going anywhere. –  Dan Jun 6 '13 at 21:52

2 Answers 2

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Hint:

We want to find the number of roots of the (holomorphic) function $h(z)=f(z)-z^n$ inside the unit circle. The main way Rouche's theorem is used in this context is by splitting up the function you're interested in into two parts, one of which is strictly less than the other on a given simple closed contour (here the boundary of the unit circle). Since $h$ is already a sum of two functions, there is a natural way to split it up...

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So you could say the original work I was attempting to do actually bears some fruitful work...I was hesitant as my proof writing abilities are not very strong –  Dan Jun 6 '13 at 21:55
    
@Dan yes, you just need to go a little further with it, by splitting $h$ up and comparing the size of the two parts on the boundary of the unit circle. –  Tom Oldfield Jun 6 '13 at 21:58
    
Both of you answered the question very well as far as I'm concerned and it was difficult to decide which to accept. On another note, I'm not a very big fan of conway as an author, what if any other text books might help my understanding of complex analysis? –  Dan Jun 7 '13 at 22:48
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I've heard that "A visual guide to complex Analysis" by Tristham Needham is good for a very intuitive approach (something which I think is important) so have bought it and an going to start reading. I also have Lang's "Complex Analysis" which covers quite a lot of material and is, I think, a staple reference. –  Tom Oldfield Jun 8 '13 at 10:26
    
I might just give that a try, I am attempting to patch the holes in my proof writing/understanding which was a product of poor instruction in the beginning and excessive workload from my other major –  Dan Jun 28 '13 at 6:22

Rouché's theorem concerns the existence of zeros. To start, let us rephrase the problem in these terms. We want find the number of zeros of $f(z)-z^n$.

In order to apply Rouché's theorem, we need a contour and two functions. Since we have an estimate for $f(z)$ along the unit circle, the natural contour to try is $|z|=1$.

Rouché's theorem tells us, once we specialize to this contour, that if $h(z)$ and $g(z)$ are analytic on the closed unit disk, and $|h(z)|<|g(z)|$ on $|z|=1$, then $g(z)$ and $g(z)+h(z)$ have the same number of zeros on the unit disk. It remains to cleverly choose $g(z)$ and $h(z)$.

It may also help to note that $|z^n|=1$ on the contour. Try to combine this with the given inequality for $f(z)$.

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