Question about complex integrals I have two questions about integrals.
1.) $$\frac{1}{2\pi i} \int_{\gamma} \frac{dw}{\sin{\frac{1}{w}}}$$ where $\gamma$ is the circle $|w| = \frac{1}{5}$.
2.) $$\int_{\gamma_{a}} \frac{z^{2} + e^{z}}{z^{2}(z-2)}dz$$
where $\gamma_{a}$ is the positively oriented circle $|z| = a, a >0, a \neq2.$
For problem $1$, I concluded that the answer is $\frac{1}{6}$ and I got this answer by computing the residue.
$$\frac{1}{\sin{\frac{1}{w}}} = \frac{1}{\frac{1}{w}} + \frac{\frac{1}{w}}{6} + \frac{7 (\frac{1}{w})^{3}}{360}+...$$
$$ \frac{1}{\sin{\frac{1}{w}}} = w + \frac{1}{6w} + \frac{7}{360w^{3}} +... $$.
But, for the second problem I'm a bit confused about how to go about solving it. If I assume that $a < 2$ then I can do this:
$$ \frac{z^{2} + e^{z}}{z^{2}(z-2)} = \frac{z^{2} +1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+...}{z^{2}(z-2)}$$
$$ \frac{z^{2} + e^{z}}{z^{2}(z-2)} = \frac{3}{2(z-2)}+\frac{1}{z^{2}(z-2)}+ \frac{1}{z(z-2)} + \frac{z}{6(z-2)} + ...$$
hence, the residue is $\frac{3}{2}$. I'm not sure if this is correct so I would appreicate someone looking over this calculation. Also, I don't know what to do when $a > 2$.
EDIT: From the suggestion in the comments, I did partial fractions on some of the terms in problem 2.
$$ \frac{1}{z^{2}(z-2)} = -\frac{1}{4z} -\frac{1}{2z^{2}} + \frac{1}{4(z-2)} $$
$$ \frac{1}{z(z-2)} = -\frac{1}{2z} =\frac{1}{2(z-2)} $$
So, the residue contributions are ($\frac{3}{2} - \frac{1}{4} + \frac{1}{4} -\frac{1}{2} +\frac{1}{2} = \frac{3}{2}.$) Is this correct? Hence, the answer would be the same as it was when $a < 2 .$

For the first problem, we enforce the transformation $z\mapsto \frac1w$ to obtain
$$\oint_{|z|=1/5}\frac1{\sin(1/w)}\,dw=\oint_{|z|=5}\frac1{\sin(z)}\frac1{z^2}\,dz\tag1$$
There are 3 poles of $\frac{1}{z^2\sin(z)}$ that are inside the circle $|z|=5$.   These include a third order pole at $z=0$ and simple poles at $z=\pm \pi$.
The associated residues are $1/6$, $-1/\pi^2$, and $-1/\pi^2$.  Putting it all together, we find that
$$\frac1{2\pi i }\oint_{|z|=1/5}\frac1{\sin(1/w)}\,dw=\frac16-\frac2{\pi^2}$$
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Alternatively, we can evaluate the integral on the left-hand side of $(1)$ directly.  The poles of $\frac1{\sin(1/w)}$ are located at $w=1/(n\pi)$ for $n\in \mathbb{Z}$.  All of these poles for $n>1$ are contained inside the circle $|w|=1/5$.
The residue of the pole at $w=1/(n\pi)$ is $(-1)^{n-1}/(n^2\pi^2)^2$.  Hence, we see that
$$\oint_{|z|=1/5}\frac1{\sin(1/w)}\,dw=2\sum_{n=2}^\infty \frac{(-1)^{n-1}}{n^2\pi^2}=\frac16-\frac2{\pi^2}$$
which agrees with the previous result.


For the second problem, there is a second order pole at $z=0$ and a simple pole at $z=2$.
The residue at $z=0$ is
$$\lim_{z\to0}\frac{d}{dz}\frac{z^2+e^z}{(z-2)}=-3/4$$
The reside at $z=2$ is $\frac{4+e^2}{4}$.
And you can finish.
