What is the biggest neighbourhood $U$ containing $3$ where $\arctan(z)$ has a convergent taylor series?
So far I have tried to use $\arctan'(z)=\frac{1}{1+z^2}$ to construct a taylor series centered at $3$ with the geometric series. It is easy to obtain a series centered at $0$ but I fail to get one at $3$. Is this even a promising approach? Or can I argue differently and a taylor series doesn't have to be constructed explicitly?
Edit: I quickly want to flesh out my idea mentioned in the comment section: Firstly, a partial fractions decomposition yields $$f'(z)=\frac{1}{1+z^{2}}=\frac{-i/2}{z-i}+\frac{i/2}{z+i}.$$ Then we have
- $\frac{1}{z-i}=-\frac{1}{i-3}\cdot\frac{1}{1-\frac{z-3}{i-3}}=\frac{1}{3-i}\cdot\sum_{n=0}^{\infty}\left(\frac{z-3}{i-3}\right)^{n}$ for $z\in D_{\sqrt{10}}(3)$.
- $\frac{1}{z+i}=\frac{1}{i+3}\cdot\frac{1}{1-\frac{z-3}{i+3}}=\frac{1}{3+i}\cdot\sum_{n=0}^{\infty}\left(\frac{z-3}{i+3}\right)^{n}$ for $z\in D_\sqrt{10}(3)$.
In total I get $$f'(z)=\sum_{n=0}^{\infty}\frac{i}{2}\left(\frac{1}{(3+i)^{n+1}}+\frac{1}{(i-3)^{n+1}}\right)(z-3)^{n}$$ and thus $$f(z)=\sum_{n=0}^{\infty}\frac{i}{2(n+1)}\left(\frac{1}{(3+i)^{n+1}}+\frac{1}{(i-3)^{n+1}}\right)(z-3)^{n+1}$$ as my taylor series converging on $D_\sqrt{10}(3)$.
If you want the biggest disk centered at $3$ such that $\arctan$ has a convergent Taylor series there, take the disk with radius $\sqrt{10}$. That's so because $\pm i$ are the singularities of $\frac1{1+z^2}$ closest to $3$ and the distance from $3$ to $\pm i$ is $\sqrt{10}$. And integrating term by term the Taylor series of $\frac1{1+z^2}$, you get the Taylor series of $\arctan$.