TITLE: Evaluating a complex function integration on a circle
QUESTION [0 upvotes]: Evaluate 
$$ \int_c \frac {1}{z \sin z} dz$$ where:
$$ C: |z| = 3 $$
I know that z = 0 is a non-removable singularity. It also might be an essential singularity since $sin z$ will cause a singular behavior if I used the Taylor expansion in the denominator. Does the function have to be analytic on the inside of the circle $ C $ to use Cauchy- Goursat theorem? Is there any way to evaluate this integral without using the residue theorem? I am trying to find a quick solution to avoid the use of the residue theorem because it is quite difficult.

REPLY [2 votes]: You can use a few summands (very few) of the power series around $\;z=0\;$ to see what's going on, and then use the well known development for a geometric series (fill in details):
$$z\sin z=z\left(z-\frac{z^3}6+\ldots\right)=z^2\left(1-\frac{z^2}6+\ldots\right)\implies$$
$$\frac1{z\sin z}=\frac1{z^2}\frac1{1-\frac{z^2}6+...}=\frac1{z^2}\left(1+\frac{z^2}6+\frac{z^4}{36}+\ldots\right)=\frac1{z^2}+\frac16+\ldots$$
and now use that
$$\oint_\gamma z^ndz=\begin{cases}2\pi i\;,&n=-1\\{}\\0\;,&-1\neq n\in\Bbb Z\end{cases}\;,\;\;\gamma\;\;\text{any simple closed path around zero}$$
Of course, the above only shows the residue is zero, which I can't see why you say it "quite difficult": you can do as shown above, or with the formula (observe zero is a double pole):
$$\lim_{z\to0} \left(z^2f(z)\right)'=\lim_{z\to0} \left(\frac z{\sin z}\right)'=\lim_{z\to0}\frac{\sin z-z\cos z}{\sin^2z}\stackrel{\text{l'Hospital}}=\lim_{z\to0}\frac{z\rlap{\;\;/}\sin z}{2\rlap{\;\;/}\sin z\cos z}=0$$
{"set_name": "stack_exchange", "score": 0, "question_id": 1766462}