The title basically says it but the question is this:
Let ord$(f, z_0) = n$ (where $z_0 \neq 0$) and let $g(z) = f(1/z)$. Show ord$(g, 1/z_0) = n$.
Given my somewhat limited understanding of poles and Laurent series, I seem to be looking to prove that inverting $f$'s input will not "interfere" with the coefficients of its Laurent series to make it so the resulting Laurent series (that of $g$) has non-zero coefficients where $f$'s did not or vice versa. From the definition I think we have that the Laurent series of $g$ about $1/z_0$ is \begin{equation*} g(z) = f\left(\frac{1}{z}\right) = \sum_{k=-\infty}^\infty a_k \left(\frac{1}{z} - \frac{1}{z_0}\right)^k \end{equation*} where the $a_k$'s are derived from the coefficients of $f$'s Laurent series in some way. However I don't know how to show that an $a_k$ will be zero iff its corresponding coefficient is zero (I did investigate somewhat into the integral form of the $a_k$'s but it proved fruitless).
I also have some basic results about the orders of poles of sums, products and reciprocals of functions but they don't seem to be relevant. Any direction with regards the solution or sources that could provide more understanding of this would be greatly appreciated.
$z_0$ is a pole of order $n$ if $\lim_{z \to z_0} (z-z_0)^n f(z) $ is finite and not zero (this is really the definition: note if $f$ is not identically zero there is only one $n$ for which this occurs). So let's look at $(z-1/z_0)^n g(z)$: $$ \lim_{z \to 1/z_0}\left( z - \frac{1}{z_0} \right)^n g(z) = \lim_{z \to 1/z_0}\left( z - \frac{1}{z_0} \right)^n f\left(\frac{1}{z}\right) = \lim_{z \to 1/z_0} \left(-\frac{z}{z_0}\right)^n\left( \frac{1}{z}-z_0 \right)^n f\left(\frac{1}{z}\right) $$ But $(-z/z_0)^n \to (-1)^n \neq 0$, so the limit is nonzero if and only if $\lim_{z \to 1/z_0} \left( \frac{1}{z}-z_0 \right)^n f\left(\frac{1}{z}\right)$ is nonzero, i.e. $f$ has a pole of order $n$ at $1/z=z_0$.