# Thread: Rouche Theorem - Proof

1. ## Rouche Theorem - Proof

This is a step in Rouche Theorem proof.
$\displaystyle f(z)$ and $\displaystyle g(z)$ are analytic on and inside a closed path $\displaystyle C$.
$\displaystyle |f(z) - g(z)| < |f(z)|$ for all $\displaystyle z$ on $\displaystyle C$

Let $\displaystyle F(z) = g(z)/f(z)$

I need to prove $\displaystyle \oint _C \frac{F'(z)}{F(z)} dz = 0$

2. Also, $\displaystyle C$ cannot pass through any zeros of f or g. Now for any function $\displaystyle f(z)$ analytic in $\displaystyle C$ except for poles:

$\displaystyle \int_{\gamma} \frac{f'(z)}{f(z)}dz=2\pi i I(f(\gamma);0)$

where $\displaystyle I(f(\gamma);0)$ is the winding number right? That's how many times the contour $\displaystyle f(\gamma)$ goes around the origin.

Now, by Rouche's theorem:

$\displaystyle |f(z)-g(z)|< |f(z)|,\quad z\in \gamma$

Since the contour $\displaystyle C$ never passes through any zero of either function, we can divide that expression by $\displaystyle |f(z)|$ and end up with:

$\displaystyle |h(\gamma)-1|<1$

That means $\displaystyle h(\gamma)$ is inside the unit circle centered at one right? What then does that say about the expression:

$\displaystyle \int_{\gamma} \frac{h'(z)}{h(z)}dz=2\pi i I(h(\gamma);0)$

3. Originally Posted by shawsend
Also, $\displaystyle C$ cannot pass through any zeros of f or g. Now for any function $\displaystyle f(z)$ analytic in $\displaystyle C$ except for poles:

$\displaystyle \int_{\gamma} \frac{f'(z)}{f(z)}dz=2\pi i I(f(\gamma);0)$

where $\displaystyle I(f(\gamma);0)$ is the winding number right? That's how many times the contour $\displaystyle f(\gamma)$ goes around the origin.

Now, by Rouche's theorem:

$\displaystyle |f(z)-g(z)|< |f(z)|,\quad z\in \gamma$

Since the contour $\displaystyle C$ never passes through any zero of either function, we can divide that expression by $\displaystyle |f(z)|$ and end up with:

$\displaystyle |h(\gamma)-1|<1$

That means $\displaystyle h(\gamma)$ is inside the unit circle centered at one right? What then does that say about the expression:

$\displaystyle \int_{\gamma} \frac{h'(z)}{h(z)}dz=2\pi i I(h(\gamma);0)$
Thanks. The inequality guarantees that the zeros don't lie on C.

Sorry but I have no clue of winding numbers etc. Isn't there a simpler explanation using maybe just Laurent Series, Cauchy Formula etc?

I follow your logic - $\displaystyle h(\gamma)$ never circles 0, hence the winding number is zero. My question would then be how do you get -

$\displaystyle \int_{\gamma} \frac{f'(z)}{f(z)}dz=2\pi i I(f(\gamma);0)$

4. Originally Posted by aman_cc
The inequality guarantees that the zeros don't lie on C.
Ok, thanks. I see that now. Don't know how to prove it without winding numbers though.

5. Thanks shawsend. I will try to read up on your first equation. I follow the steps after that.

6. Originally Posted by aman_cc
My question would then be how do you get -

$\displaystyle \int_{\gamma} \frac{f'(z)}{f(z)}dz=2\pi i I(f(\gamma);0)$
Let me try. I need to understand it better too.

So the winding number is just the number of times a contour goes around a point in the complex plane. For the contour $\displaystyle \gamma$ around the origin, we can express it as:

$\displaystyle I(\gamma;0)=\frac{1}{2\pi i}\oint_{\gamma} \frac{dz}{z}$

and for the contour traced by $\displaystyle f(\gamma)$, we could write:

$\displaystyle I(f(\gamma);0)=\frac{1}{2\pi i}\int_{f(\gamma)}\frac{1}{w}dw$

but if $\displaystyle w=f(\gamma(t))$ and $\displaystyle \gamma:[a,b]\to \mathbb{C}\;$ then:

$\displaystyle I(f(\gamma);0)=\frac{1}{2\pi i}\int_{f(\gamma)}\frac{1}{w}dw=\frac{1}{2\pi i}\int_{a}^{b}\frac{f'(\gamma(t))\gamma'(t)dt}{f(\ gamma(t))}$

$\displaystyle =\frac{1}{2\pi i}\int_{\gamma} \frac{f'(z)}{f(z)}dz$