Laurent serie of $ \frac { \cos z}{ \sin z + \sinh z - 2z}$ I'm working on an example given in my book of complex analysis:
$$ \frac { \cos z}{ \sin z + \sinh z - 2z}$$
but I can't figure out how he finded the residue in 0.
The few steps he is showing make me think he has computed the first few terms of its Laurent serie (like here : Laurent series $\frac{1}{\sin(z)}$ around $z=0$).
Nevertheless, I can't find the same result. Am I doing something wrong ?
So can you help find the laurent serie, and do you have another method to find the residue ?

I found that 
$$ \frac 1 { \sin z + \sinh z - 2z} = \frac{ 5! } {2} z^{-5} - \frac{(5!)^2}{2 \cdot 9!} z^{-1}  + o(z^2) $$
The book gives :
$$ \frac {\cos z} { \sin z + \sinh z - 2z} =
\frac{ 5! } {2} z^{-5} (1 - z^2 /2 + 125/3024 z^4 +...) $$
The residue is equal to $ \frac {625 i \pi} {126} $

You have$$\sin(z)+\sinh(z)-2z=\frac{z^5}{60}+\frac{z^9}{181\,440}+\cdots=z^5\left(\frac1{60}+\frac{z^4}{181\,440}+\cdots\right),$$where the $\cdots$ only has terms whose exponent is a multiple of $4$. So, you should compute$$\frac{\cos z}{\frac1{60}+\frac{z^4}{181\,440}+\cdots}=a_0+a_2z^2+a_4z^4+\cdots$$This means that you have\begin{align}1-\frac{z^2}2+\frac{z^4}{24}+\cdots&=\left(\frac1{60}+\frac{z^4}{181\,440}+\cdots\right)\left(a_0+a_2z^2+a_4z^4+\cdots\right)\\&=\frac{a_0}{60}+\frac{a_2}{60}z^2+\left(\frac{a_4}{60}+\frac{a_0}{181\,440}\right)z^4+\cdots\end{align}So, you know that $a_0=60$ and now you get $a_4$ by solving the equation$$\frac{a_4}{60}+\frac{a_0}{181\,440}=\frac1{24}.$$So,$$\frac{\cos z}{\sin(z)+\sinh(z)-2z}=\frac{a_0}{z^5}+\frac{a_4}z+\cdots$$and the residue that you're after is $a_4=\dfrac{625}{252}$.
