# Why is the Taylor expansion of $\frac{1}{\sin(z)}$ not this?

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

$$\int_{C(0,1)}\frac{1}{z^2\sin(z)}$$dz

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 $$\frac{1}{z}+\frac{z}{3!}....$$

• 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 $\sin z$). Oct 4, 2022 at 15:31
• For starters, is $\dfrac 1{2+3} = \dfrac12 + \dfrac13$? Oct 4, 2022 at 15:31
• First of all, these are Laurent series, not Taylor series. Secondly, the main tool you will need is the geometric series $$\frac 1{1-u} = 1+u+u^2+\dots \quad\text{when } |u|<1.$$ Oct 4, 2022 at 15:37
• 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?
– uoiu
Oct 4, 2022 at 15:38
• @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? Oct 4, 2022 at 16:00

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.