I seem to have taken some bad lecture notes about this...

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 $$\oint_{|z| = \frac{\pi}{2}} \frac{1}{\sin(z)}\ dz$$


  • $\begingroup$ Sorry my mistake, everything including the $a_{-3}$ term and before are all equal zero, corrected. $\endgroup$ – tgun926 Jun 13 '13 at 8:55
  • $\begingroup$ @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... $\endgroup$ – 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 $z$.

  • $\begingroup$ I don't get this. Why does this imply that everything before $a_{-1}$ are 0? $\endgroup$ – tgun926 Jun 13 '13 at 9:35
  • $\begingroup$ 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?) $\endgroup$ – Gerry Myerson Jun 13 '13 at 9:38
  • $\begingroup$ 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 $\endgroup$ – tgun926 Jun 13 '13 at 9:49
  • $\begingroup$ $g(z)=a_0+a_1z+a_2z^2+\dots$ where $a_0\ne0$. You should think about why this implies that $1/g(z)=b_0+b_1z+b_2z^2+\dots$. $\endgroup$ – Gerry Myerson Jun 13 '13 at 12:24

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