$$\int_{\gamma}\frac{z}{(z-1)^{2014}}dz$$, where $\gamma$ is the diamond with vertices $-2, -2i, 2, 2i $.

Question

kirbykirby · Answer

I tried using Cauchy's Integral formula:
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_{\gamma}\frac{f(z)}{{(z-z_0)}^{n+1}}dz$$
So I let $f(z)=z$ thus giving $f^{(n)}(z)=0$ and so $f^{(n)}(-1)=0$
Which implies that the integral must be 0?

Although I find it strange because there's the point z=1 inside the closed curve where it's not analytic, and thus this would not satisfy Cauchy's Theorem?

anonymous · Answer

$\color{blue}{\text{Originally Posted by}}$ @SithsAndGiggles
I'm pretty sure that just because a function isn't analytic on a region with a closed curve $C$, this doesn't mean that the integral is *not* necessarily zero.
$\color{blue}{\text{End of Quote}}$

anonymous · Answer

$\color{blue}{\text{Originally Posted by}}$ @SithsAndGiggles
So just because it doesn't satisfy Cauchy's theorem does not mean that $\int_C f(z)~dz\not=0$.
$\color{blue}{\text{End of Quote}}$

kirbykirby · Answer

Ok good to know!