Can someone point out where the mistake is please.

Let $\gamma = C(0,1)$ with $\gamma : [0,1] \to \mathbb{C}$. Okay so it is know that $\int_{\gamma}\frac{dz}{z} = 2\pi i$. And that the winding number $n(\gamma,0) = 1$. However lets try to calculate this from the definition.

$$2\pi i n(\gamma,0)=\int_{\gamma}\frac{dz}{z}=\int_0^1\frac{\gamma\prime(t)dt}{\gamma(t)} = \text{Log}(\gamma(1))-\text{Log}(\gamma(0))$$ $$=\text{ln}|\gamma(1)|-\text{ln}|\gamma(0)| + i(\theta(\gamma(1)) - \theta(\gamma(0))).$$ But we have that $\gamma(0)=\gamma(1)$. So is the above expression not just zero? Meaning that this winding number should be zero for any $\gamma$ closed curve. Clearly this is wrong but I do not understand why.


To use the fundamental theorem of calculus, you need an anti-derivative on the whole contour. But your "Log" is not such a thing. The complex logarithm is more complicated than that. See https://en.wikipedia.org/wiki/Complex_logarithm

  • $\begingroup$ Okay, but what if we just shift the whole picture to the right by 2 units so that nothing intersects the negative real axis, doesn't the same thing happen? $\endgroup$ – pureundergrad Oct 19 '18 at 21:39
  • $\begingroup$ There is a legitimate antiderivative for $1/z$ in the half plane $\mathrm{Re}\;z > 0$. $\endgroup$ – GEdgar Oct 19 '18 at 21:44
  • $\begingroup$ @pureundergrad, $1/z$ has no singularity at $2$, so the integral is $0$. The point is exactly that there is singularity inside contour around $0$. $\endgroup$ – Ennar Oct 19 '18 at 21:44

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