$$\int_\gamma \frac{\cos z}{(z-z_0)^3}dz,\quad \gamma \colon |z|=3$$
I was able to find the solution for this problem which included $z=2 \in \gamma, -3<z<3$, therefore $-3<2<3$. And when comparing to Cauchy formula we find that $n+1=3 \Rightarrow n=2$.
Now the rest of the solution is obvious. What I wasn't able to understand was how did we determine that $<=2 \in \gamma$ and that it is the point that we are looking for.
Please note that English isn't my first language so I'm having problems translating the exact mathematical words to English so pardon the quality of this post. Any help would be greatly appreciated.
Assume that $z_0 \notin \overline{D(0,3)}$ thus exists $s>0$ such that $|z_0| \geq 3+s$
Then $f(z)=\frac{\cos{z}}{(z-z_0)^3}$ is holomorphic on the disc $D(0,3+\frac{s}{2}) \supset \gamma$ thus from the theorem of Cachy in convex regions we have that $\int_{\gamma}f(z)dz=0$
Now if $|z_0|<3$ we have from Cauchy's the integral formula for derivatives for $g(z)=\cos{z}$ we have that $$2 \pi i g''(z_0)=\int_{\gamma}\frac{\cos{z_0}}{(z-z_0)^3}dz$$