I am trying to solve a problem where I am asked to compute the Taylor expansion series of the function $\sin\left(\frac{1}{1-z}\right)$ around $z=0$. Now, I know that to find the coefficients of the series I can use the Cauchy Integral formula, but here is my doubt: the Taylor series for $\sin(z)$ is valid when $z$ is closed to $0$, and in my case when $z\sim 0$ I have $\sin\left(\frac{1}{1-z}\right) \sim \sin(1)$.
Does everything work the same? Why?
On
You can use Bell polynomials and the Taylor series of $\frac{1}{1-z}$ and $\sin z$, so we have : $$\frac{1}{1-z}=\sum_{n=0}^\infty z^n,\ \sin z=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ Then we have : $$\sin \biggl(\frac{1}{1-z}\biggr)=\sum_{n=0}^\infty\frac{\sum_{k=0}^n \alpha_kB_{n,k}(0!,1!,...,(n-k+1)!)}{n!}z^n$$ Where : $$\alpha_{2n}=0,\\ \alpha_{2n+1}=(-1)^n\frac{1}{(2n+1)!}$$ For more information see the link https://en.wikipedia.org/wiki/Bell_polynomials.
On
By the Faa di Bruno formula and some properties for the Bell polynomials of the second kind, we obtain \begin{align*} \biggl[\sin\biggl(\frac{1}{1-z}\biggr)\biggr]^{(n)} &=\sum_{k=0}^{n}\sin^{(k)}\biggl(\frac{1}{1-z}\biggr)B_{n,k}\biggl(\frac{1!}{(1-z)^2},\frac{2!}{(1-z)^3},\dotsc, \frac{(n-k+1)!}{(1-z)^{n-k+2}}\biggr)\\ &=\sum_{k=0}^{n}\sin\biggl(\frac{1}{1-z}+\frac{k\pi}{2}\biggr) \frac{1}{(1-z)^{n+k}} B_{n,k}(1!,2!,\dotsc,(n-k+1)!)\\ &=\sum_{k=0}^{n}\sin\biggl(\frac{1}{1-z}+\frac{k\pi}{2}\biggr) \frac{1}{(1-z)^{n+k}} \binom{n-1}{k-1}\frac{n!}{k!}\\ &\to n!\sum_{k=0}^{n}\sin\biggl(1+\frac{k\pi}{2}\biggr) \binom{n-1}{k-1}\frac{1}{k!} \end{align*} as $z\to0$, where $B_{n,k}(x_1,x_2,\dotsc, x_{n-k+1})$ denotes the Bell polynomials of the second kind. Consequently, we acquire \begin{equation*} \sin\biggl(\frac{1}{1-z}\biggr)=\sum_{n=0}^{\infty}\Biggl[\sum_{k=0}^{n}\sin\biggl(1+\frac{k\pi}{2}\biggr) \binom{n-1}{k-1}\frac{1}{k!}\Biggr]z^n, \quad z\in[-1,1). \end{equation*} Note that the Faa di Bruno formula, the Bell polynomials of the second kind, and their proprties and identies can be found in the following references.
References
$\sin\left(\frac{1}{1-z}\right)=\sin\left(1+\frac{z}{1-z}\right)=\sin 1 \cos \left(\frac{z}{1-z}\right) + \cos 1 \sin \left(\frac{z}{1-z}\right),$ so just use Taylor expansions of $\sin$ and $\cos$ around zero for $y=\frac{z}{1-z} \sim 0$.