<s>[INST] Find the Taylor and Laurent series with center at $z_{o}= 1$ of the function $\frac{\sinh(z)}{(z-1)^4}$ Let $$f(z)=\frac{\sinh(z)}{(z-1)^4}.$$
First I do the following:
$$\sinh(z)=\sinh((z-1)+1)=\sinh(z-1)\cosh(1)+\cosh(z-1)\sinh(1)$$
The expansions of $\sinh(z)$ and $\cosh(z)$ are
$$\begin{align*} 
\sinh(z) &= z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots \\
\cosh(z) &= 1+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots 
\end{align*}$$
Then
$$\begin{align*} 
\sinh(z-1) &= (z-1)+\frac{(z-1)^3}{3!}+\frac{(z-1)^5}{5!}+\cdots \\
\cosh(z-1) &= 1+\frac{(z-1)^2}{2!}+\frac{(z-1)^4}{4!}+\cdots
\end{align*}$$
Now let's see what
$$\begin{align*} 
\frac{\cosh(1)\sinh(z-1)}{(z-1)^4} 
&= \cosh(1) \biggl( \frac{1}{(z-1)^3}+\frac{1}{3!(z-1)}+\frac{(z-1)}{5!}+\cdots \biggr) \\
\frac{\sinh(1)\cosh(z-1)}{(z-1)^4} 
&= \sinh(1) \biggl( \frac{1}{(z-1)^4}+\frac{1}{2!(z-1)^2}+\frac{1}{4!}+\cdots \biggr)
\end{align*}$$
Rewriting these last two equalities we have
$$\begin{align*} 
\frac{\cosh(1)\sinh(z-1)}{(z-1)^4} 
&= \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} \\
\frac{\sinh(1)\cosh(z-1)}{(z-1)^4} 
&= \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4}
\end{align*}$$
Hence the Taylor and Laurent series of $f(z)$ is
$$
f(z) = \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} + \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4}
$$
Where we have
$$
a_{n} = \frac{\cosh(1)}{(2n-1)!} 
\quad \text{and} \quad 
b_{n} = \frac{\sinh(1)}{(2n)!}.
$$
But I don't know if it's okay. I feel a bit confused regarding these series. Could you tell me if I'm okay? or in another case, could you give me any suggestions?
 [/INST] Your work is good but you can make life a bit easier starting with $z=t+1$ (this is what you implicitly did)
$$\frac{\sinh (z)}{(z-1)^4}=\frac{\sinh (t+1)}{t^4}$$ Expanding as you did
$$\sinh (t+1)=\sinh (1) \cosh (t)+\cosh (1) \sinh (t)$$ Now, using the expansions of $\cosh (t)$ and $\sinh (t)$ around $t=0$ you then have
$$\frac{\sinh (t+1)}{t^4}=\sum_{n=0}^\infty a_n\, t^{n-4}$$ where
$$a_{2n+1}=\frac {\cosh(1)}{(2n+1)!}\qquad \text{and} \qquad a_{2n}=\frac {\sinh(1)}{(2n)!}$$ which is your result  (just make $t=z-1$).
</s>