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**davesface** Use the identity $\displaystyle sinh(z+ \pi i)=-sinh(z)$ and the fact that $\displaystyle sinh(z)$ has a period of $\displaystyle 2\pi i$ to find the Taylor series for $\displaystyle sinh(z)$ about $\displaystyle z_0=\pi i$.

Clearly the numerator will be $\displaystyle (z-z_0)^{2n+1}=(z-\pi i)^{2n+1}$ because of the periodicity, and the answer overall is going to be negative because of the given identity, but I'm not sure what the denominator is going to be here or how to justify it other than just saying that it's obvious from the definition of a Taylor series.