Main Question
A very common form of Cauchy's integral theorem is
Let $A\subset\mathbf{C}$ be simply connected and open, $f:A\rightarrow\mathbf{C}$ analytic, $I$ a compact interval, and $\gamma:I\rightarrow A$ continuous (or whatever) with $\gamma(I)\subset A$. Then $\int_\gamma f(z)dz=0$.
I think it would also suffice (and not make the proof any harder) to assume that $\gamma$ is homotopic in $A$ to a point without requiring that $A$ is simply connected. Am I correct?
If so, why is it never (or seldom) stated this way? The same applies to the residue theorem and other related theorems of complex analysis.
Related
I would be less surprised if the following statement was completely obvious:
Let $A\subset\mathbf{C}$ open, $I$ a compact interval, and $\gamma:I\rightarrow A$ a continuous path homotopic in $A$ to a point. Then there exists $U\subset A$ open and simply connected with $\gamma(I)\subset U$.
I suspect this is true, and I have a vague idea how to prove it. But it's not a one-liner and I don't remember seeing it in any textbook. Am I missing something?
Best Answer
Let's try this curve, where $A$ is the complement of $\{0\}$.
Certainly $\int_\gamma\frac{dz}{z} = 0$, where $\gamma$ is that contour.