1. ## sinh Taylor Series

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

Clearly the numerator will be $(z-z_0)^{2n+1}=(z-\pi i)^{2n+1}$ 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.

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

Clearly the numerator will be $(z-z_0)^{2n+1}=(z-\pi i)^{2n+1}$ 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.
Do you know the Taylor series for sinh(z) about $z_0= 0$?

3. It's just the $2n+1$ terms of the $e^z$ expansion, which leads me to think that the denominator would just be $(2n+1)!$, although it seems like you could derive that without knowing anything about the periodicity.

Thanks.