Taylor series and singularities

Question

I'm studying complex analysis and I cannot find a way to solve this exercise. I'll report the text: Set $$f(z)=\frac{1}{1-\sin(z)}$$ 1 - Find the first nonzero terms in Taylor expansion of $f(z)$ around $0$ $$$$2 - Find and classify the isolated singularities in the extended complex plane $$$$3 - Evaluate the integral $$\int_{\left | z=2 \right |} \frac{f(z))}{z}dz$$

Now, I was able to solve the first part of the exercise by computing the Taylor series by definition and I found the isolated singularities of the functions (all points such that $z = \frac{\pi}{2}+2k\pi$, with $k$ integer. By computing the limit of the function in the neighbourhood of the singularities, I'm able to find out that those singularities are poles, but I'm not able to find the order. From this point on I'm stuck. For part 3, I know I should use the residue theorem to compute the integral by knowing that the function has two singularities at $0$ and $\frac{\pi}{2}$. Anyone knows how to solve it? Thanks in advance

Answer 1

Using what Fred said I was able to complete the second part of the question. In order to compute the other part, it is needed to proceed from the definition which will lead to an infinite series sum and obtain that the result is $\frac{-8}{\pi^2}$ The help of my complex analysis professor was needed though to check the result.

Answer 2

Let $g(z):=1- \sin z$ and $z_k:=\frac{\pi}{2}+2k\pi$

Then we have

$$g(z_k)=0, g'(z_k)=0, g''(z_k) \ne 0.$$

Hence $g$ has azero of order $2$ at $z_k.$ Therefore $f$ has a pole of order $2$ at $z_k.$

Can you proceed ?