In our lecture, we hat the following Cauchy-integral-theorem (CIT):
Let $f$ be a function that is holomorphic in a neighbourhood of the disk $\mathcal K(z_0,M)$ with radius M and centre $z_0$. Then $$f(z_0)=\frac{1}{2\pi i}\int_{\gamma(z_0,M)}\frac{f(z)}{z-z_0}\ \text{d}z$$ $\underline{Note}:$ $\gamma(z_0,M)$ is the boundary of the disk with radius $M$ and centre $z_0$
Now I saw some examples of the CIT like this:
Calculate $\int_{|z|=2}\frac{\sin(z)}{z+i}\ \text{d}z$
For $z_0=-i$ and $f(z)=\sin(z)$ we obtain with the CIT $$\int_{|z|=2}\frac{\sin(z)}{z+i}\ \text{d}z = 2\pi if(-i)=2\pi i\sin(-i) $$
What I don't understand here is that the disk in the integral to be calculated is $|z|=2 \Leftrightarrow \gamma(0,2)$, so $z_0$ actually would need to be $0$. I don't see why the example is consistent with the CIF. If the integral would be $\int_{\gamma(-i,2)}\frac{\sin(z)}{z+i}\ \text{d}z$ instead of $\int_{\gamma(0,2)}\frac{\sin(z)}{z+i}\ \text{d}z$, everything would be clear.
Thank you!
$f(z_0)=\frac{1}{2\pi i}\int_{\gamma(z_0,M)}\frac{f(z)}{z-z_0}\ \text{d}z$
is only a special case of the $CIT$.
More general we have
$f(w)=\frac{1}{2\pi i}\int_{\gamma(z_0,M)}\frac{f(z)}{z-w}\ \text{d}z$ for all $w$ such that $|w-z_0|<M$.