Find principal part of Laurent series

Question

Find principal part of Laurent series:

$$f(z)= \frac{1}{\sin z + \sinh z - 2z}$$

I calculate it and I have something like this:

$$\frac{1}{\dfrac{2z^5}{5!}+\dfrac{2z^9}{9!}+\cdots}$$ and don't know what to do next...

Answer 1

Just following the comments, it is clear that we just need the Taylor series of: $$ f(z)=\frac{1}{\sum_{n\geq 0}\frac{2 z^{4n}}{(4n+5)!}}=\sum_{n\geq 0}a_n z^n\tag{1}$$ till the term $a_4 z^4$. By evaluating the LHS and the RHS at $z=0$ we get $a_0=60$, and since the LHS is $g(z^4)$ for some analytic function $g$, we have $a_1=a_2=a_3=0$. Moreover: $$ \frac{1}{\sum_{n\geq 0}\frac{2 z^{4n}}{(4n+5)!}} = \frac{1}{\left(\frac{2}{5!}+\frac{2z^4}{9!}\right)\left(1+O(z^8)\right)}=\frac{1}{\frac{2}{5!}+\frac{2z^4}{9!}}\left(1+O(z^8)\right)\tag{2}$$ so: $$f(z)=60\cdot\frac{1}{1+\frac{z^4}{3024}}\left(1+O(z^8)\right)=60\left(1-\frac{z^4}{3024}\right)\left(1+O(z^8)\right)\tag{3} $$ leads to $\color{red}{a_4=-\frac{5}{252}}$.