Show that the function $f(z) = \frac{\sin(z)}{z^n}$ has an anti-derivative $F(z)$ in $\mathbb{C}\setminus \{0\} \iff n$ odd.

I think that you have to show that $f(z)$ is analytic in $\mathbb{C}\setminus \{0\} \iff n$ odd, and tried writing $\sin(z) = z - \frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}$ and then dividing by$z^n$, and it's clear that then the series when $n$ is even contains odd powers of $z$, suggesting divergence and thus non-analyticy. And for $n$ odd, the series only contains even powers, but I can't prove it exactly. Can anybody help?

• The issue is whether you have a $z^{-1}$ term before you try to take an antiderivative. If $n$ is not, you can verify that such a term doesn't exist, and then you can check that the Laurent series converges. However, if there is such a term, then you can argue that if your function had an analytic anti-derivative, then $\log$ would have to be an analytic function on $\mathbb C \backslash\{0\}$ – Aaron Jan 2 '18 at 18:36
• Also for $n=0$. – lhf Jan 2 '18 at 18:39
• Ah I see, yes, that's actually very simple! Thank you. I understand your argument. – Dan Jan 2 '18 at 18:39

Let $n=2k+1$, then $$f(z)=\frac{1}{z^{2k+1}}\sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)!}\\ = \sum_{n=0}^\infty \frac{z^{2(n-k)}}{(2n+1)!}$$ Now, the first few of these terms may well be negative powers of $z$ (if not, this function is a power series, and so is analytic, and so has an antiderivative), depending on how big $k$ is, but what happens when you integrate something like $$\frac{1}{z^{2j}}$$ in a closed loop around $0$? Note that the way we construct antiderivatives is by defining $$F(z)=\int_0^1f(\gamma(t))\gamma'(t)\mathrm dt$$ where $\gamma$ is some path from some point (this will determine the constant up to which antiderivatives are defined) to the point $z$ in $\mathbb{C}\setminus \{ 0\}$. Note that by our computation, the choice of path $\gamma$ doesn't matter, and so this is a well defined function.