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**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:

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