Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: Please let me know if I need to fix anything.

share|cite|improve this question
up vote 1 down vote accepted

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.


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.

share|cite|improve this answer
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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.