Trouble with determining principal part of function at a pole

Question

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.

Answer 1

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.

Answer 2

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