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.

Question

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?

Answer 1

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.