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I am trying to understand this proof of Rouché's theorem, but I am missing the logic of the last and most crucial step. Here are the assumptions:

Suppose that $ f $ and $ g $ are analytic inside and on a regular closed curve $ \gamma $ and that $ |f(z)| \gt |g(z)| $ for all $ z > \in \gamma $. Then $$ \mathcal{Z}(f + g) = \mathcal{Z}(f) \text{ inside > } \gamma $$

The proof goes as follows:

First note that $ f \neq 0$ on $ \gamma $ since otherwise $ |g| \lt 0 $ which doesn't make any sense. So, we can write the following:

$$ \mathcal{Z}(f + g) = \frac{1}{2\pi i} \int_{\gamma}\frac{(f+g)'}{(f+g)} = \frac{1}{2\pi i} \int_{\gamma} \frac{(f(1 + \frac{g}{f}))'}{f(1 + \frac{g}{f})} = \frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} + \frac{1}{2\pi i} \int_{\gamma} \frac{(1 + \frac{g}{f})'}{(1 + \frac{g}{f})} $$

We know $ f $ is analytic, so $ \frac{1}{2\pi i} \int_{\gamma} \frac{f'}{f} = \mathcal{Z}(f) $ by the argument principle. What I don't know is, why is $ \frac{1}{2\pi i} \int_{\gamma} \frac{(1 + \frac{g}{f})'}{(1 + \frac{g}{f})} = 0$?

Jean-Claude Arbaut
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Max
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Since $\bigl|{g(z)\over f(z)}\bigr|<1$, it follows that $1+{g(z)\over f(z)}$ must lie in the right half plane for all $z\in \gamma$. But then the curve $z\mapsto 1+{g(z)\over f(z)}$ cannot encircle $0$, so $Z(1+{g(z)\over f(z)})=0$.

Per Erik Manne
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  • That makes sense. But why is $ P(1 + \frac{g}{f}) $ also 0? – Max May 11 '13 at 09:16
  • What is the definition of $P$? If it is the integral in the question, you can use the argument principle, which you have quoted already. – Per Erik Manne May 11 '13 at 09:19
  • I rephrased the question so it doesn't use Z or P(the number of poles). You can look here: http://math.stackexchange.com/questions/388308/why-does-frac12-pi-i-int-gamma-fracfzfz-dz-eta-f-circ – Max May 11 '13 at 09:40
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The key thing to note is that $\frac{|g|}{|f|} \in (0,1)$ and so the image of $\gamma$ under the function $1+\frac{g}{f}$ lies strictly inside the open ball of radius one about the point $1$ in the complex plane.

Drawing a picture should convince yourself that the winding number about the origin of any curve with this property is zero. Formally, the principal branch of lograrithm is a continuous choice of argument for this curve, and so the winding number is zero from that.

Note that the argument principle tells us that this means that the number of zeroes of $1+\frac{g}{f}$ equals the number of poles of $1+ \frac{g}{f}$ inside $\gamma$, but it does not tell us how many of each there are.

A slightly different proof merely shows that the winding number of $h=\frac{f+g}{f}$ is zero using the above argument, and then noting that since $f$ and $g$ are holomorphic inside $\gamma$, poles of $h$ correspond to zeroes of $f$. Thus since $\mathcal{Z}(h)=\mathcal{P}(h)$ the number of zeroes of $f+g$ equals the number of zeroes of $f$, which is what we wanted. Of course, all of this is only true when counting with multiplicity.

Tom Oldfield
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  • I just don't see why the winding number is relevant. Clearly it is, but why I am using the winding number around 0 to solve this integral? – Max May 11 '13 at 09:38
  • I actually rephrased the question, so perhaps you could look at that: http://math.stackexchange.com/questions/388308/why-does-frac12-pi-i-int-gamma-fracfzfz-dz-eta-f-circ – Max May 11 '13 at 09:39
  • @Max In this case, the integral is the definition of the winding number about zero of the image of $\gamma$, so finding the winding number evaluates the integral. – Tom Oldfield May 11 '13 at 09:42
  • @Max the linked question is different, although the result is made use of in my answer (and any proof I can think of of Rouche's theorem.) – Tom Oldfield May 11 '13 at 09:52