Find the Laurent series of an indeterminate function

Question

$$ f(z) = \frac{z}{(\sin z)^2} $$ at $z_0 = 0$ (for the first four terms).

So I thought I knew what to do, but I don't. Since it appears to be an indeterminte form, could I by L'hopital turn it into:

$$\frac{1}{2(\sin z)(\cos z)} $$ but even here I am jammed

Answer 1

Why just the first four terms? Let's compute all of them. For first, since $z=0$ is a removable singularity for $\frac{\sin z}{z}$ we have that $z=0$ is a simple pole for $f(z)=\frac{z}{\sin^2 z}$, and the problem boils down to finding the Taylor series, in a neighbourhood of the origin, of: $$ g(z) = \left(\frac{z}{\sin z}\right)^2. \tag{1}$$ From the Weierstrass product: $$\frac{\sin z}{z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2\pi^2}\right)\tag{2} $$ we have that any $z=m\pi$, for $m\in\mathbb{Z}\setminus\{0\}$, is a simple pole for $\frac{z}{\sin z}$, and by considering the logarithmic derivative of $(2)$ we get: $$ \cot z-\frac{1}{z} = \sum_{n\geq 1}\frac{2z}{z^2-n^2\pi^2},\tag{3} $$ a really useful identity, that leads to: $$ \frac{1-z\cot z}{2}=\sum_{n\geq 1}\frac{z^2}{n^2\pi^2-z^2}=\sum_{n\geq 1}\frac{\frac{z^2}{n^2\pi^2}}{1-\frac{z^2}{n^2\pi^2}}=\frac{\zeta(2)}{\pi^2}z^2+\frac{\zeta(4)}{\pi^4}z^4+\ldots\tag{4} $$ Rearranging we have: $$\cot z = \frac{1}{z}-2\sum_{n\geq 1}\frac{\zeta(2n)}{\pi^{2n}}z^{2n-1}\tag{5}$$ and now we just need to differentiate both terms with respect to $z$ and multiply them by $-z^2$ to get:

$$\frac{z^2}{\sin^2 z}=1+2\sum_{n\geq 1}\frac{\zeta(2n)(2n-1)}{\pi^{2n}}z^{2n}\tag{6}$$

from which: $$ g(z)=\left(\frac{z}{\sin z}\right)^2 = 1+\frac{z^2}{3}+\frac{z^4}{15}+o(z^5) $$ readily follows.

Answer 2

Hint:
The function $f(z) = \frac{1}{\sin^2z}$ has a pole of order $2$ at $z=0$ therefore its Laurent series around $z=0$ is of the form $\frac{a_{-2}}{z^2}+\frac{a_{-1}}{z}+a_0+a_1z+\dots$
Since $\sin^2z = z^2-\frac{z^4}{3}+\frac{2z^6}{45}+\dots$ we have in a punctured disc around $z=0$: $$ \Big( \frac{a_{-2}}{z^2}+\frac{a_{-1}}{z}+a_0+a_1z+\dots \Big)\Big( z^2-\frac{z^4}{3}+\frac{2z^6}{45}+\dots \Big) = 1 $$

You can calculate the terms you need by comparing coefficients.

Answer 3

This is a Laurent series, so we would expect that the limit does not exist if we have a pole: that is what is happening here. To get a characterisation of it, the simplest thing to do is work out the series expansion directly.

First, expand $\sin{z}$ as a Taylor series about $z=0$: $$ \sin{z} = z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040}+O(z^9) $$ (this should be plenty of terms... you can't necessarily tell at this point.) Now, we insert this series into the denominator of your function: $$ \frac{z}{(z-z^3/6+z^5/120+z^7/5040+O(z^9))^2}. $$ Now, we can pull out a factor of $z^2$ in the denominator to get $$ \frac{1}{z} \left( 1-\frac{z^2}{6} + \frac{z^4}{120} -\frac{z^6}{5040}+O(z^8) \right)^{-2}. $$

At this point we use the binomial expansion on the bracket, to get $$ \frac{1}{z} \left( 1-2\left(-\frac{z^2}{6} + \frac{z^4}{120}-\frac{z^6}{5040}+O(z^8)\right)+\frac{2\cdot 4}{2!}\left(-\frac{z^2}{6} + \frac{z^4}{120} +O(z^6)\right)^2 -\frac{2\cdot4\cdot6}{3!}\left( \frac{z^2}{6} \right)^3 +O(z^6) \right), $$ where I have thrown away terms that will not count in the final reckoning because we can essentially see now that only terms up to $z^5$ will appear. Multiplying out the brackets and keeping the relevant terms will give you the answer.