Calculate $\int_{\left|z-1\right|=2}z^{n}\sin\left(z\right)dz$ for $n\in \mathbb{Z}$

Question

Calculate $$\int_{\left|z-1\right|=2}z^{n}\sin\left(z\right)dz$$ for $n\in \mathbb{Z}$

My attempt: According to the following result which was presented at my course as Cauchy's integral formula for Disks

Let $U\subseteq \mathbb{C}$ open, $z_{0}\in U$, $f:U\rightarrow\mathbb{C}$ holomorphic, $R>0$ such that $\overline{\mathcal{B}_{R}(z_{0})}\subseteq U$, then $$\int_{\left|z-z_{0}\right|=R}f\left(z\right)dz=0$$

We have $$\int_{\left|z-1\right|=2}z^{n}\sin\left(z\right)dz=0$$ for $n\geq 0$.

For another hand, according to the following result which was presented at my course as *Cauchy's integral formula, in general, *

Let $U\subseteq \mathbb{C}$ open, $z_{0}\in U$, $f:U\rightarrow\mathbb{C}$ holomorphic, $r>0$ such that $\overline{\mathcal{B}_{r}(z_{0})}\subseteq U$ and $a\in \mathcal{B}_{r}(z_{0})$, then $$f(a)=\frac{1}{2\pi i}\int_{\left|z-z_{0}\right|=r}\frac{f\left(z\right)}{z-a}dz=0$$

We can extract the following corollary

Let $U\subseteq \mathbb{C}$ open, $z_{0}\in U$, $f:U\rightarrow\mathbb{C}$ holomorphic, $r>0$ such that $\overline{\mathcal{B}_{r}(z_{0})}\subseteq U$ and $a\in \mathcal{B}_{r}(z_{0})$, then $$f^{(n)}(a)=\frac{n!}{2\pi i}\int_{\left|z-z_{0}\right|=r}\frac{f\left(z\right)}{(z-a)^{n+1}}dz$$

Considering $f(z)=\sin (z)$, $r=2$ and $a=0$, we have $$\int_{\left|z-1\right|=2}z^{-n}\sin\left(z\right)dz=\frac{2\pi i}{(n-1)!} \sin \left(\frac{(n-1) \pi}{2}\right)$$ for $n> 0$. This completes the case that we were missing.

My question: Is my reasoning correct? Have I used the theorems well?

Answer 1

Yes, the argument is fine but you might want to remark where the $\sin ((n-1) \pi /2)$ comes from. I had to look twice to get this. Moreover, you might want to state that you apply the formula with $n-1$ at the end.

Another way to do this would be using Laurent series and the residue theorem.

The residue theorem you get that the integral is $2 \pi i $ times the residue of $z^n \sin z$ at $0$. (Everywhere else the residue is clearly zero.)

Now, you know that $z^{n} \sin(z)= z^n\sum_{k = 0}^{\infty} \frac{1}{(2k+1)!}z^{2k+1} = \sum_{k = 0}^{\infty} \frac{(-1)^k}{(2k+1)!}z^{2k+1+n}$ and the residue is the coefficient of $z^{-1}$.

This is clearly $0$ for $n\ge 0$ and for $n$ odd, and for the negative even ones you'd get $\frac{(-1)^{1+n/2}}{(n-1)!}$ as you did.