Trouble with determining principal part of function at a pole

In Fischer's $\textit{A Course in Complex Analysis}$ I am encountering some difficulty in explicitly calculating the principal part of a function at a pole. The function is $f(z)= \frac{1}{z - \sin z}$ with isolated singularity $z_0 = 0$. I eventually figured out that $0$ is a pole of order $3$. So in the principal part, one of the terms I should get is $6/z^3$, correct? Here is how I determined that:

Let $g(z)= z^3 f(z)$. Then we have that $$\begin{array} {lcl} \lim_{z \rightarrow 0} g(z)&=& \lim_{z \rightarrow 0} \frac{z^3}{z- \sin z} \\ &=& \lim_{z \rightarrow 0} \frac{z^3}{z -(z- \frac {z^3}{3!}+ \frac{z^5}{5!}- \frac{z^7}{7!}+ \cdots)} \\&=& \lim_{z \rightarrow 0} \frac{z^3}{\frac{z^3}{3!}- \frac{z^5}{5!}+ \frac{z^7}{7!}- \cdots}\\&=& \lim_{z \rightarrow 0} \frac{1}{\frac{1}{3!}- \frac{z^2}{5!}+ \frac{z^4}{7!}- \cdots} \\&=& 3!=6. \end{array}$$
Since this limit is not $0$ but is finite, we have a pole of order $3$ at $0$ for $f$. Because $f$ is holomorphic on a punctured neighborhood of $0$, $g$, which is bounded in, say, a punctured open disk $D_r(0)-\{0\}$ about $0$ with some radius $r>0$ and also holomorphic there, has a unique holomorphic extension $\hat{g}$ on the open disk $D_r(0)$ via the Riemann extension theorem. What I tried to do next was to somehow write $f$ as a power series in terms of $\hat{g}$ by letting $\hat{g}(z)= a_0 + a_1 z+ a_2 z^2+ \cdots$ with $z \in D_r(0)- \{0\}$, but I don't think that is actually of any help in determining the principal part of $f$ at $0$. I believe that $6/z^3$ might just be a portion of the principal part at that particular singularity, and according to WolframAlpha, $6/z^3 + 3/10z$ comprise the negative powers of the power series expansion of $f$ at $z=0$. I got this from the following reference: http://www.wolframalpha.com/input/?i=1%2F%28z+-+sin%28z%29%29 Please let me know if I need to fix anything.

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I think that you already did it: $$\frac{1}{z-\sin z} = z^{-3}\frac{1}{\frac{1}{3!}- \frac{z^2}{5!}+ \frac{z^4}{7!}- \cdots}$$

Now, you just need to invert a power series with a non-zero constant term.

Spoiler:

Based on the wiki, $b_0 = 1/a_0 = {3!} = 6$ and $b_2 = -b_0a_2/a_0 = 6\cdot 1/{5!}\cdot {3!}=3/10$. So you would get: $$\frac{1}{\frac{1}{3!}- \frac{z^2}{5!}+ \cdots} = 6 + \frac{3}{10}z^2+\cdots$$ Then one can multiply by $z^{-3}$ both sides.

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The Laurent series expansion can help in determining the type of isolated singularity, correct? For instance, would the isolated singularity $z_0= 0$ be a pole of the function $(\sin z+ \cos z -1)^{-2}$? I think it would also be of order $2$. –  Libertron Feb 2 '14 at 0:03

Try finding the Laurent series of $$\frac{1}{z-\sin(z)}$$ first then take the limits

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I tried doing that, but the $1$ in the numerator is giving me the problem. How must this be properly factorized? –  Libertron Feb 1 '14 at 0:11
Previously, you expanded $\sin(z)$ as a series. What you should have done is expand $\frac{1}{z-\sin(z)}$ instead.. The first 2 terms are $\frac{6}{x^3} +\frac{3}{10x}+\cdots$, and I'll leave you to compute the rest. You could do it in a way similar to computing the Maclaurin series if you wish –  Millardo Peacecraft Feb 1 '14 at 0:18
Do you mean a Taylor series about $z=0$? OK, that is fine, but how am I certain that the negative powers will be accounted for? –  Libertron Feb 1 '14 at 0:25