# Laurent series of $\sin z/(1 - \cos z)$

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?

• Does this help? wolframalpha.com/input/?i=sin(z)%2F(1-cos(z)) Look under Series expansion at z=0 (Laurent series) Commented Apr 10, 2019 at 20:54
• Unfortunately no because I need a step by step solution. It is an exercise of an exam. Commented Apr 10, 2019 at 20:56

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

• 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)? Commented Apr 10, 2019 at 21:24
• 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). Commented Apr 10, 2019 at 21:27
• Shouldn't you also prove that |yl less than 1 while using that expansion? Commented May 4, 2020 at 22:08

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, $$\dfrac{\sin(z)}{1-\cos(z)}= \dfrac{d}{dz}[\log(1-\cos(z))].$$

Applying @Nelson Faustino's first step result, as well as a fact that $$\cot (x)=\sum _{n=0}^{\infty } \frac{(-1)^n 2^{2 n} B_{2 n} x^{2 n-1}}{(2 n)!}$$ where $$B_{2n}$$ is Bernoulli numbers, one can obtain that $$\frac{\sin z}{1-\cos z}=2\sum _{n=0}^{\infty } \frac{(-1)^n B_{2 n} z^{2 n-1}}{(2 n)!}=\frac{2}{z}-\frac{z}{6}-\frac{z^3}{360}+O(z^3)$$