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I have trouble solving this exercise: find the first three terms of the Laurent series of $\sin z/(1 - \cos z)$ centered at $z=0$.

I have found the first two. I proved that at $z=0$ we have a first order pole and the first one I calculated the residue. I also thought that the second term is zero because this function is odd. Now I have problems with the third. Someone can help me?

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You know that laurent series of $sin(z) = z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7})$.

Then, the laurent series of $ 1-cos(z)= \frac{z^2}{2}-\frac{z^4}{24}+O(z^{6})$

Overall you have $\frac{z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7})}{\frac{z^2}{2}-\frac{z^4}{24}+O(z^{6})}$.

Now look at the denominator: $(\frac{z^2}{2}-\frac{z^4}{24}+\frac{z^6}{720}+O(z^8))^{-1}=(\frac{z^2}{2})^{-1}(1-\frac{z^2}{12}+\frac{z^4}{360}+O(z^{6}))^{-1}=\frac{2}{z^2}(1-(-\frac{z^2}{12}+\frac{z^4}{360})+\frac{z^4}{144}+O(z^6)) =\frac{2}{z^2}(1+\frac{z^2}{12}+\frac{z^4}{240})=\frac{2}{z^2}+\frac{1}{6}+\frac{z^2}{120}$

Multiply everything together to get: $(z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7}))$ $(\frac{2}{z^2}+\frac{1}{6}+\frac{z^2}{120} )$ $ =\frac{2}{z}-\frac{z}{6}-\frac{z^3}{360} +O(z^5).$

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  • $\begingroup$ I don't understand why you eliminate the exponent "-1" of the denominator. Have you used the first order expansion of (1+x)^a that is (1+ax)? $\endgroup$ – Antonio Sannia Apr 10 at 21:24
  • $\begingroup$ Yes, it's the series expansion of $(1+y)^{-1}$, but I am using 2 terms here, ie $1-y+y^2+....$ (and ignoring al powers of z greater than 6). $\endgroup$ – Alexandros Apr 10 at 21:27
  • $\begingroup$ Perfect, thank you! $\endgroup$ – Antonio Sannia Apr 10 at 21:38
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First, we recall that the identities $\sin(z)=2\sin(z/2)\cos(z/2)$ and $2\sin^2(z/2)=1-\cos(z)$ allows us to show that $$ \dfrac{\sin(z)}{1-\cos(z)}=\dfrac{2\sin(z/2)\cos(z/2)}{2\sin^2(z/2)}=\cot(z/2). $$

Thus, the Laurent series of $\dfrac{\sin(z)}{1-\cos(z)}$ is essentially the Laurent series of $\cot(z/2)=i\dfrac{1+e^{-iz}}{1-e^{-iz}}$.

On the other hand, we notice in case where $e^{\Im(z)}=|e^{-iz}|<1$, that is $\Im(z)<0$, one can employ the geometric series expansion $$ \dfrac{1}{1-e^{-iz}}=\sum_{k=0}^\infty e^{-ikz} $$ to show that $$ \sum_{k=0}^\infty i (e^{-ikz}-e^{-i(k+1)z}) $$ is a (possible) Laurent series expansion.

Remark 1: One can also use the fact that $\cot(z/2)$ is a meromorphic function with poles $z=2k\pi$ ($k\in \mathbb{Z}$) to find an alternative Laurent series expansion by means of residue theory.

Remark 2: Other possibility may be considered using the fact that the $\dfrac{\sin(z)}{1-\cos(z)}$ corresponds to the logarithmic derivative of $1-\cos(z)$. Indeed, $$ \dfrac{\sin(z)}{1-\cos(z)}= \dfrac{d}{dz}[\log(1-\cos(z))].$$

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