# Laurent series expansion of 1/sin(z) - why is everything before residue 0?

I have written (about $$z = 0$$)

$$\frac{1}{\sin(z)} = \ldots +\frac{a_{-3}}{z^3}+\frac{a_{-1}}{z^{-1}}+{a_{1}}{z^1}+{a_{3}}{z^3}+\ldots$$

Mentioned $$\sin(z)$$ being an odd function, and everything including and before the term $$a_{-3}$$ is $$0$$. I don't understand that bit. Also, why isn't $$a_{-1}=0$$ as well?

The next part is to evaluate $$\oint_{|z| = \frac{\pi}{2}} \frac{1}{\sin(z)}\ dz$$

Thanks

• Sorry my mistake, everything including the $a_{-3}$ term and before are all equal zero, corrected. – tgun926 Jun 13 '13 at 8:55
• @user14111 I never long divided $\sin(z)$... The next step is multiply both sides by $\sin(z)$, so that you get $1 = (res + a_1z + a_3z^3 ..)(z - z^3/3! + z^5/5!...)$, then compaire coefficients... – tgun926 Jun 13 '13 at 9:33

Another approach: define

$$f(z):=\frac1{\sin z}\implies \lim_{z\to 0}\,zf(z)=\lim_{z\to 0}\frac z{\sin z}=1$$

Which means $\,z=0\,$ is a pole of order $\;1\;$ of $\,f(z)\,$ , and this means that in the Laurent series around zero for this function we get

$$a_n=0\;\;\forall\,n\in\Bbb Z\;,\;n<-1\;\;\wedge\;\;f(z)= \frac 1z+a_0+a_1z+\ldots$$

$\sin z=zg(z)$ where $g(0)\ne0$, so $1/\sin z=(1/z)h(z)$ where $h(z)$ has an expansion in nonnegative powers of $z$.

• I don't get this. Why does this imply that everything before $a_{-1}$ are 0? – tgun926 Jun 13 '13 at 9:35
• What happens when you take $h(z)$, which has an expansion in nonnegative powers of $z$, and multiply it by $1/z$? What powers of $z$ do you get? (or, rather, what powers of $z$ are impossible to get?) – Gerry Myerson Jun 13 '13 at 9:38
• Powers less than $z^-1$ are impossible... But what I don't get is how you come up with $h(z)$ in the first place. If you say $\sin z = zg(z)$, doesn't that mean $g(z)$ is the taylor series expansion with even powers instead? How does this correlate to $h(z)$ as being the expansion of nonnegative powers? Thanks – tgun926 Jun 13 '13 at 9:49
• $g(z)=a_0+a_1z+a_2z^2+\dots$ where $a_0\ne0$. You should think about why this implies that $1/g(z)=b_0+b_1z+b_2z^2+\dots$. – Gerry Myerson Jun 13 '13 at 12:24