I'm wondering how is the Maclaurin series of $\tan z$ where $z\in\mathbb{C}$. Some first idea is see it: \begin{eqnarray} \tan z = \frac{\sin z}{\cos z} \end{eqnarray} and the Maclaurin series respectively are: \begin{eqnarray} \sin z = \sum_{n=0}^{\infty} \frac{(-1)^{n}z^{2n+1}}{(2n+1)!} \hspace{0.5cm} \text{and }\hspace{0.5cm} \cos z = \sum_{n=0}^{\infty} \frac{(-1)^{n}z^{2n}}{(2n)!} \end{eqnarray} but the quotient seems weird. Do you know other way to find this series?
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1See https://math.stackexchange.com/questions/1805660/formula-for-the-general-term-of-the-taylor-series-of-tanx-at-x-0 – Henry Oct 05 '21 at 00:11
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There is a very good StackExchange post here https://math.stackexchange.com/a/2099213/254075 which is much superior to the answer this question was closed in favor of. – sharding4 Oct 05 '21 at 01:55
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@sharding4: I've added the corresponding question to the "already has answers here" list. (I don't know what the reputation threshold is for making such additions. If you see an "Edit" item in the top-right corner of the box, then you have the power.) – Blue Oct 05 '21 at 07:16