If $f(z)$ is analytic in $ 0 < |z| < n$, what is the residue of the function $f(z^2)$ at $z = 0$?
Attempt If $f(z)$ has a pole of order n at $z=0$ it seems like the residue of $f(z^2)$ would just be the coefficient of the $a_{-n}$ term of the Laurent series for $f(z)$, since the Laurent series for $$ f(z) = \frac{a_{-n}}{z^{n}} + \cdots + \frac{a_{-2}}{z^{2}} + \frac{a_{-1}}{z} + a_{0} + a_{1}z + a_{2}z^2 + \cdots $$ and $z^{n^{2}} = z^{2n}$ is this right?
You should check your definition of Residue, is not the $a_{-n}$ coefficient in the Laurent expansion but the $a_{-1}$ one.
If $f$ has a pole of order $n$ at $z=0$ then $$ f(z)=a_{-n}z^{-n} + \cdots + a_{-1}z^{-1} + \sum_{k=0}^{\infty}a_kz^k $$ with $a_{-n} \neq 0$. Of course $Res(f,0)=a_{-1}$. Then around $0$, we have that if $g(z)=f(z^2)$ then \begin{align} g(z)=f(z^2) & =a_{-n}z^{-2n} + \cdots + a_{-1}z^{-2} + \sum_{k=0}^{\infty}a_kz^{2k}\\ & = b_{-2n}z^{-2n} + \cdots + b_{-1}z^{-1} + \sum_{k=0}^{\infty}b_kz^{k}\\ \end{align} where $$ b_j=\left\{ \begin{array}{cc} a_{j/2} & \text{ if } j \in 2 \mathbb{Z} \\ 0 & \text{in other case} \end{array} \right. $$ Since $b_{-2n}=a_{-n} \neq 0$ then $g$ has a pole of order $2n$ at $z=0$ and $Res(g,0)=b_{-1}=0$.
Now is clear that this still holds even if $f$ has not a finite order pole, since the residue of $g$ at $z=0$ will still be $b_{-1}=0$. Thus supposing that $f$ has a pole of order $n$ at $0$ only helps to get that $g$ has a pole of order $2n$ at $0$.