# Winding number always zero by definition. Help find the mistake.

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.

## 1 Answer

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

• 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? – pureundergrad Oct 19 '18 at 21:39
• There is a legitimate antiderivative for $1/z$ in the half plane $\mathrm{Re}\;z > 0$. – GEdgar Oct 19 '18 at 21:44
• @pureundergrad, $1/z$ has no singularity at $2$, so the integral is $0$. The point is exactly that there is singularity inside contour around $0$. – Ennar Oct 19 '18 at 21:44