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Let $f(z)$ be an elliptic function, $f(z + 1) = f(z + \tau) = f(z)$. If $f(z)$ has only simple poles $z_j$, then there is an expansion using the Weierstrass $\zeta$ functions, $$ f(z) = C(\tau) + \sum_j\Big[\operatorname{Res}_{z=z_j}f(z) \Big]\zeta(z - z_j|\tau) $$ where $C(τ)$ is a constant depending on $\tau$. When $f(z)$ has higher order poles, a similar expansion exists using $\zeta^{(m)}(z - z_j)$.

However, I wonder if similar expansion is possible for $f(z)^{1/n}$, for some integer $n$?

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  • $\begingroup$ For $f(z)^{1/n}$ to exist as a meromorphic function, its roots and poles must have degree divisible by $n$, no? I imagine after dividing those degrees by $n$, then, you can just do the higher order pole expansion using $\zeta^{(m)}$ you mentioned. $\endgroup$ Commented Aug 26 at 17:24
  • $\begingroup$ I guess my problem will be when the pole orders of $f(z)$ is not divisible by $n$, and $f(z)^{1/n}$ has some branch cuts. $\endgroup$ Commented Aug 27 at 3:38
  • $\begingroup$ Then $f(z)^{1/n}$ would not be an elliptic function. $\endgroup$ Commented Aug 27 at 21:03

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