I'm learning complex analysis, specifically Laurent series, and need help with the following exercise:
If $f(z)$ is entire and for some $k \in \mathbb{N}$ the function $g(z) = z^kf(\frac{1}{z})$ is bounded in some annular region $\Delta(0, 0, \delta)$ with $\delta > 0$, show that $f(z)$ is a polynomial.
Note: the annulus $\Delta(0, 0, \delta)$ is defined as the region where $0 < |z| < \delta$.
Here's my work so far:
Since $f(z)$ is entire, by Taylor's theorem, we can expand $f(z)$ as a power series centered at $z_0 = 0$, and so
$$f(z) = \sum_{n = 0}^{\infty}a_n z^n \implies f\left(\frac{1}{z}\right) = \sum_{n = 0}^{\infty}a_n \frac{1}{z^n}.$$
Therefore
$$g(z) = z^kf(\frac{1}{z}) = a_0z^k + a_1z^{k-1} + a_2z^{k-2} + \ldots + a_{k-1}z + a_k + \frac{a_{k+1}}{z} + \frac{a_{k+2}}{z^2} + \ldots$$
This is as far as I got. I think the fact that $g(z)$ is bounded is key here but I don't know how to make use of this assumption for $g(z)$.
Edit: Considering the answer of Eric Wofsey, we have
$$g(z) = \sum_{n = 1}^{\infty}\frac{a_{n+k}}{z^n} + a_k + a_{k-1}z + \ldots + a_1z^{k-1} + a_0z^k.$$
However, since $z_0 = 0$ is a removable singularity of $g(z)$, by uniqueness of Laurent series, $g(z)$ is actually the Taylor series
$$g(z) = a_k + a_{k-1}z + \ldots + a_1z^{k-1} + a_0z^k.$$
It follows that
$$f(1/z) = \frac{g(z)}{z^k} = a_0 + \frac{a_1}{z} + \ldots + \frac{a_{k-1}}{z^{k-1}} + \frac{a_k}{z^k}$$
and so
$$f(z) = a_0 + a_1z + \ldots + a_{k-1}z^{k-1} + a_kz^k$$
which shows that $f(z)$ is a polynomial of degree $k$.
If $g(z)$ is bounded on $\Delta(0,0,\delta)$, then its singularity at $0$ is removable, so it can be extended analytically to $0$. This means that $g(z)$ has a Taylor series around $0$, and so by uniqueness of Laurent series your Laurent series for $g(z)$ must actually be a Taylor series. You should be able to finish from here.