Find the poles and residue of the function $\frac{1}{z-\sin z }$ at $z=0$? $$\frac{1}{z-\sin z}$$
$$\frac{1}{z(1-\frac{\sin z}{z})}$$
the laurent's transform is also used to simplify the function.
the residue is found a $\frac{2*36*2}{5}$
but the actual answer is given as $\frac{3}{10}$
the residue is given as
The solution is incorrect and also the pole is of the order 3.
 A: One may write, by using a Taylor series expansion of $\sin z$, as $z \to 0$
$$
\begin{align}
\frac{1}{z-\sin z}&=\frac{1}{z-(z-\frac{z^3}{6}+\frac{z^5}{120}+O(z^7))}
\\\\&=\frac6{z^3}\cdot \frac{1}{1-\frac{z^2}{20}+O(z^4)}
\\\\&=\frac6{z^3}\cdot \left(1+\frac{z^2}{20}+O(z^4)\right)
\\\\&=\frac6{z^3}+\frac{6}{20z}+O(z)
\\\\&=\frac6{z^3}+\frac{3}{10z}+O(z)
\end{align}
$$ giving the expected answer.
A: Use the Taylor Series for the $\sin z$ function:
$$\sin z \approx z - \frac{z^3}{6} + \frac{z^5}{5!} + \ldots$$
In this way your denominator becomes
$$\frac{1}{\frac{z^3}{6} - \frac{z^5}{5!} + \ldots}$$
Collect a $z^3/6$ term:
$$\frac{1}{\frac{z^3}{6}\left(1 - \frac{z^2}{20}\right)} = \frac{6}{z^3} \cdot \frac{1}{1 - \frac{z^2}{20}}$$
Now the term $\frac{1}{1 - \frac{z^2}{20}}$  can be expanded with the Geometric Series, as long as $|z^2/20| < 1$:
$$\frac{6}{z^3} \sum_{k = 0}^{+\infty} \frac{z^{2k}}{20^k}$$
And the first terms of the series are
$$\frac{6}{z^3} + \frac{6}{20 z} + \frac{6 z^4}{400} + \ldots$$
And you now know that the residue is nothing but the coefficient of the $z^{-1}$ terms, which is
$$\frac{3}{10}$$
Also you can see the pole is order three, since you have a $z^{-3}$ factor in the series.
