Im struggling with complex analysis integrals, more specifically this one:


My solution so far is:

  1. Taylor expansion of $\frac{1}{\sin(z)}$, which I do by follow method which returns me the wrong result and I don't know why this wouldn't work.

    I know Taylorexpansion for $\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\mathcal{O}(z^7)$

    Which makes me think that Taylorexpansion of $\frac{1}{\sin(z)}$ is simply the same but inverted, that is $\frac{1}{z}-\frac{3!}{z^3}+\frac{5!}{z^5}+\mathcal{O}(\frac{1}{z^7})$ which is not the case. Can someone explain why this does not work?

The correct Taylorexpansion according to the book is $\frac{1}{z}+\frac{z}{3!}....$

  • 8
    $\begingroup$ It is not true that $\frac{1}{a+b} = \frac{1}{a}+\frac{1}{b}$, why do you think that it would be true for an infinite sum (like the Taylor series for $\sin z$). $\endgroup$ Oct 4, 2022 at 15:31
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    $\begingroup$ For starters, is $\dfrac 1{2+3} = \dfrac12 + \dfrac13$? $\endgroup$ Oct 4, 2022 at 15:31
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    $\begingroup$ First of all, these are Laurent series, not Taylor series. Secondly, the main tool you will need is the geometric series $$\frac 1{1-u} = 1+u+u^2+\dots \quad\text{when } |u|<1.$$ $\endgroup$ Oct 4, 2022 at 15:37
  • $\begingroup$ Please explain the first comment, I don't understand and the only thing I've done is reversing the sin(z) expansion? Cant see that I've broken any rules? $\endgroup$
    – uoiu
    Oct 4, 2022 at 15:38
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    $\begingroup$ @zzz__ You have concluded that since $\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\mathcal{O}(z^7)$, it follows that $$\frac{1}{\sin(z)} = \frac{1}{z-\frac{z^3}{3!}+\frac{z^5}{5!}+\mathcal{O}(z^7)} = \frac{1}{z} - \frac{1}{z^3/3!} + \frac{1}{z^5/5!} + \frac{1}{\mathcal{O}(z^7)} = \frac{1}{z} - \frac{3!}{z^3} + \frac{5!}{z^5} + \frac{1}{\mathcal{O}(z^7)}. $$ Do not see the error? $\endgroup$
    – Xander Henderson
    Oct 4, 2022 at 16:00

2 Answers 2


In a small neighborhood of zero, you can write: $$\dfrac 1{\sin z}=\dfrac 1{z-\dfrac {z^3}{3!}+\mathcal O(z^5)}=\dfrac 1z\cdot \dfrac 1{1-\dfrac {z^2}{3!}+\mathcal O(z^4)}=\dfrac 1z(\sum_{n\ge0}(\dfrac {z^2}{3!})^n=\dfrac 1z+\dfrac z6+\dfrac {z^3}{(3!)^2}+\mathcal O(z^5)$$, for $\lvert z\rvert \lt\sqrt6$.

But you certainly don't just invert the terms of the Taylor series for $\sin z$, as commented.

Note also that this is a Laurent series (not a Taylor series).


Write the denominator as $$z^3(1-z^2/6++...).$$You need to find, in the Laurent series, the coefficient of $z^{-1}$ so you need the coefficient of $z^2$ in the reciprocal of $1-z^2/6+...$ which is 1/6.


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