Taylor series expansion of $(e^{\pi z}-1)^2$ about $2i$?

Question

How do you get the Taylor series expansion of $(e^{\pi z}-1)^2$ about $2i$? Or in general, how can I get the Taylor series expansion of $f(x)^k$?

Answer 1

$(\exp{\pi z} - 1)^2 = \exp{2\pi z} -2\exp{\pi z} + 1$. Let $\xi = z - 2i$, then we have the following:

\begin{align*} \exp{2\pi z} -2\exp{\pi z} + 1 &= \exp{2\pi \xi} - 2\exp{\pi \xi} + 1\\ &= \sum_{n \geq 0} \frac{(2\pi\xi)^n}{n!} - 2\sum_{n \geq 0} \frac{(\pi\xi)^n}{n!} + 1\\ &= \sum_{n \geq 1} \frac{(2^n - 1)\pi^n}{n!} \xi^n\\ &= \sum_{n \geq 1} \frac{(2^n - 1)\pi^n}{n!} (z - 2i)^n\\ \end{align*}

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In the general case $f(z)^k$ with $f \in \mathcal{O}(U)$, where $U \subset_{\text{ open}} \mathbb{C}$ and $k \in \mathbb{N}$, it's less evident. If $k$ small, you can consider expanding as above. Nonetheless, this could get pretty tedious for large values of $k$. If $f$ has a series expansion at $z_0 \in U$:

\begin{align*} f(z) = \sum_{n \geq 0} a_n(z-z_0)^n, \text{for all z in some nhb of } z_0 \end{align*}

Then absolute convergence yields us the following well-known Cauchy product:

\begin{align*} f(z)^k &= (\sum_{n \geq 0} a_n(z-z_0)^n)^k, \text{for all z in some nhb of } z_0 \\ &= \sum_{m \geq 0} \Big(\sum_{j_1 + \dots + j_k = m} a_{j_1}a_{j_2}...a_{j_k}\Big) (z-z_0)^m \end{align*}

Therefore it amounts to getting a closed form of the summation $\sum_{j_1 + \dots + j_k = m} a_{j_1}a_{j_2}...a_{j_k}$, which is sometimes easy with a little bit of intuition.

Answer 2

Hint: Observe \begin{align} (e^{\pi z}-1)^2 = e^{2\pi z}-2e^{\pi z} +1. \end{align} Next, observe \begin{align} e^{\pi z} = e^{\pi z -2\pi i} = e^{\pi(z-2i)} = \sum^\infty_{n=0} \frac{\pi^n(z-2i)^n}{n!}. \end{align}