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My teacher wrote the above equation on the board the other day and acted like it was obvious. Here, $ \eta $ is the winding number of $ f \circ \gamma $. Here is the justification he gave:

$$ \frac{1}{2\pi i} \int_{\gamma} \frac{f'(z)}{f(z)} dz = \frac{1}{2\pi i} \int_{a}^{b} \frac{f'(z(t))}{f(z(t))} z'(t)\ dt = \int_{f \circ \gamma}\frac{1}{\zeta} d\zeta = \eta( f \circ \gamma, 0) $$

I think he may be missing a $ \frac{1}{2\pi i} $, since the last equality would follow from the Cauchy Residue Theorem if a $ \frac{1}{2\pi i} $ was placed in front of the integral. What I really don't understand is this equation:

$$ \frac{1}{2\pi i} \int_{a}^{b} \frac{f'(z(t))}{f(z(t))} z'(t)\ dt = \int_{f \circ \gamma}\frac{1}{\zeta} d\zeta $$

Help?

Max
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  • You're right, the $\frac 1{2i\pi}$ is missing, since it wasn't "absorbed" by the change of variable. You found the error yourself and corrected it! Good job. – Philippe Malot May 11 '13 at 09:41

1 Answers1

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This last part comes from parametrising the curve $f\circ\gamma$. If $\gamma :[a,b]\rightarrow \mathbb{C}$ is a parametrisation of $\gamma$, then $f\circ\gamma;[a,b]\rightarrow \mathbb{C}$ is a parameterisation of the image of $\gamma$ under $f$. Thus the integral

$$\int_{f \circ \gamma}\frac{1}{\zeta} d\zeta=\int_{f \circ \gamma}\frac{1}{z} dz$$

can be evaluated in the usual way by

$$\int_a^b\frac{1}{f\circ\gamma(t)}(f\circ\gamma)'(t)dt=\int_a^b\frac{f'(\gamma(t))}{f(\gamma(t))}\gamma'(t)$$

The missing $2\pi i$ is an error, but the above is the real work being done in that substitution.

Tom Oldfield
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