2
$\begingroup$

I met an integral like this: $\oint_{|z|=1}\frac{dz}{z^2 sin(z)}$. I know that it can be worked out by using Residue theorem. How can we calculate it using other methods, such as Cauchy integral formula? (In fact, this integral is an exercise in a chapter before the Residue theorem in a book on complex analysis. )

$\endgroup$

1 Answer 1

3
$\begingroup$

HINT:

Write $\frac{1}{z^2\sin(z)}=\frac{z\csc(z)}{z^3}$ and note that $f(z)=z\csc(z)=\frac{z}{\sin(z)}$ is holomorphic on $|z|\le 1$ (with a removeable singularity at $z=0$).

Then, apply Cauchy's Integral Formula

$$f''(0)=\frac{2!}{2\pi i}\oint_{|z|=1}\frac{f(z)}{z^3}\,dz$$

with $f(z)=\frac{z}{\sin(z)}$.


NOTE: Calculating $f''(0)$ will take a bit of work but is facilitated by using series expansion. The result is $f''(0)=\frac13$.

$\endgroup$
4
  • $\begingroup$ Thanks, Dr. MV. But I still have a little confusion on holomorphic function. We know that a complex function f(z) is said to be holomorphic at a point z0, if it is differentiable at every point within some open disk centered at z0. So I do not know why $f(z)=\frac{z}{sin(z)}$ is still holomorphic due to the existence of a removable singularity. $\endgroup$
    – Mark_Phys
    Mar 15, 2017 at 5:38
  • $\begingroup$ Note that $1/f(z)=\frac{\sin(z)}{z}=\sum_{n=0}^\infty\frac{(-1)^nz^{2n}}{(2n+1)!}$ which is holomorphic in the plane. $\endgroup$
    – Mark Viola
    Mar 15, 2017 at 5:50
  • $\begingroup$ You mean "reciprocal of holomorphic function" is also holomorphic? Is this correct? $\endgroup$
    – Mark_Phys
    Mar 15, 2017 at 5:54
  • $\begingroup$ Well, provided that the function $f$ has no zeros, which is the case here for $|z|\le 1$. $\endgroup$
    – Mark Viola
    Mar 15, 2017 at 5:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .