# Why can't I apply Cauchy's integral theorem with the function 1/z?

From Cauchy's integral theorem we get that if $$f:U\rightarrow \mathbb{C}$$ is an holomorphic funcion then $$\oint_{\gamma}f=0$$ now, if I define a funcion $$f:\mathbb{C}\setminus\{0\}\rightarrow \mathbb{C}$$ so that $$f(z)=\frac{1}{z}$$, then we have that $$f$$ is holomorphic for all $$z$$ in the domain. Let $$\gamma(t) = e^{it},t \text{ in } [0,2\pi]$$. Because of Cauchy's integral theorem $$\oint_{\gamma}\frac{1}{z}\text{ d}z=0$$ But when I evaluated the integral I got $$2\pi i$$. Why can't I apply Cauchy's integral theorem?

• You left out the part about $U$ being simply connected. Commented Apr 13, 2020 at 16:59
• More generally, it's important to pay attention to the whole statement of any theorem. Many people concentrate too much on formulas and ignore prose. Commented Apr 13, 2020 at 17:53
• For any theorem of the form "if P then Q", if you can't verify that P is true, then you can't apply the theorem. Commented Apr 13, 2020 at 18:20

To apply Cauchy's Theorem you need $$\gamma$$ to be null-homotopic (homotopic to a point curve $$\gamma_p(t) \equiv p \in U$$) in $$U$$.
In your example this isn't the case: $$\gamma$$ is null-homotopic in $$\Bbb C$$ but not in $$\Bbb C \setminus \{0\}$$ (you can't continously "contract" the curve to a single point because there is a hole in your set)
You need to pay attention to your set $$U$$.
The problem here is the path $$\gamma(t) = e^{it}, t\in[0,2\pi)$$. It describes a full circle around the origin and the origin happens to be a singularity (more precisely a pole) of the function $$f(z) = z^{-1}$$. Usually, one defines the winding number or index of a path as $$ind_{\gamma}(z) := \frac{1}{2\pi i} \int_{\gamma} \frac{1}{\zeta-z}d\zeta$$ Now you call a path homologous to zero if for every $$z\in\mathbb{C}\backslash U$$ the index vanishes. The only point in $$\mathbb{C}$$ that is not in $$U$$ is $$z=0$$ so if you plug that in you get $$ind_{\gamma}(0) = \frac{1}{2\pi i}\int_0^{2\pi} \frac{1}{e^{i\zeta}}(e^{i\zeta})'d \zeta = \frac{1}{2\pi i}\int_0^{2\pi} \frac{i}{e^{i\zeta}}e^{i\zeta}d \zeta = \frac{i}{2\pi i}\int_0^{2\pi} d \zeta = 1$$ (funnily that is just the integral you are trying to calculate) so you see that Cauchy's Theorem cannot hold. The problem is that paths around points where the function is not holomorphic poses a problem. In fact, Morera's Theorem somewhat gives a converse to Cauchy's theorem which states that a continuous function on a domain $$U\subseteq \mathbb{C}$$ which satisfies $$\int_{\gamma} f(z) dz=0$$ for every closed piecewise $$C^1$$ path in $$U$$ must be holomorphic on $$U$$.