Let $ \gamma $ be the unit circle then $ \int_\gamma \frac {dz}{z^2 − 2z} = -\pi i$

Question

Definition: If $ f $ is holomorpic in $G$ and gamma $\gamma$ is $G$-homotophic to a point then gamma is G-contratible and if gamma is G-contractible then $ \int_\gamma f = 0 $.

By splitting the integral using partial fraction we get $$ \int_\gamma \frac {dz}{z^2 − 2z} = \frac12\int_\gamma \frac {dz}{z − 2}-\frac12\int_\gamma \frac {dz}{z} $$ Thus if we apply the definition mentioned above to the first integral on the right hand-side it will be equal to zero since $\frac 1{z-2}$ is holomorphic in $\Bbb C$ \ {2} and $ \gamma $ is $\Bbb C$ \ {2}-contractible.

However, and this is my question, by the same reasoning the second integral in the right hand-side should be zero too since $ \frac1z $ is holomorphic in $\Bbb C$ \ {0} and $ \gamma $ is $\Bbb C$-contratible (not sure), I know that it is evaluated to $-\pi i$ but what is it restricting the function for the definition to be applied in that integral?

Answer 1

If both $0$ and $2$ are in the interior of $\gamma$, by the residue theorem we have:

$$\begin{align} \int_\gamma \frac{{\rm d}z}{z^2 - 2z} &= 2\pi i\left({\rm Res}(f,0) + {\rm Res}(f,2)\right) \\ &= 2\pi i\left(\frac{1}{2\cdot 0 - 2} + \frac{1}{2\cdot 2 - 2}\right) \\ &= 2\pi i\left(-\frac{1}{2} + \frac{1}{2}\right) \\ &= 0 \end{align}$$

If $\gamma$ has $0$, but not $2$, in its interior, then $\gamma$ is $\Bbb C \setminus \{2\}$-contractible and since $1/(z-2)$ is holomorphic there, your first integral in the RHS will be zero, so the overall result will be $-\pi i$. On the other hand, if $\gamma$ has only $2$, and not $0$, in its interior, then your second integral on the RHS will be zero, because $\gamma$ is $\Bbb C\setminus \{0\}$-contractible and $1/z$ is holomorphic there, so the overall integral evaluates to $\pi i$.

If neither $0$ nor $2$ is in the interior of $\gamma$, the argument above goes for both integrals and all of it evaluates to zero.

Finally, if both point are in the interior of $\gamma$, the integral gives zero because the residues cancel each other, and this is not related to homotopy theory or.whatever - this is a particularity of that specific $f$. We've seen that at the beginning of the answer, I checked it with the residue theorem, even.

This covers all cases.

Answer 2

A possible source of confusion can come from the following wrong reasoning: we want to calculate the integral of $f(z)=1/z$ along the unit circle $\gamma$. If we do the inversion of the complex plane $w=1/z$ then $\gamma\mapsto\gamma$ and our function becomes $f(1/z)=w$ which is analytic in the unit disc, hence the integral is zero. In other words, $\gamma$ is contractable on the Riemann sphere via $\infty$ and the function $1/z$ is analytical at $\infty$.

The problem with this reasoning is the following: when we prove that the integral is zero using contraction to a finite point, it is important that the integrand at the point is bounded and the path length goes to zero, thus "bounded times zero is zero". When we do contraction via infinity, $1/z$ does go to zero, however, the path becomes infinitely long, so we get $0\cdot\infty$ which can be whatever.

Formally, when we do the variable change $w=\frac{1}{z}$ we get $$ \frac{1}{z}\,dz=w\cdot \left(-\frac{1}{w^2}\right)\,dw=-\frac{1}{w}\,dw $$ i.e. still having trouble at $w=0$. However, if the denominator degree is at least two more than that of the numerator (to compensate $1/w^2$), we can use this kind of thinking.