Using Cauchy's integral theorem without polynomial in denominator I'm trying to find $$\oint_C \frac{z^2+1}{e^{\frac {z}{10}}-1}dz$$
where $C$ is the unit circle traversed counterclockwise. I think I should use Cauchy's integral theorem but I don't understand how because the theorem specifically has a polynomial in the denominator. The integrand has a singular point at $z=0$ so I can use the Residue theorem but I don't understand how to compute the residue in this case either.
 A: I'm not sure how Cauchy's theorem would be applied here, but in the case of computing the residue, note that $$\lim_{z \to0}(z-0)\frac{z^2+1}{e^{\frac {z}{10}}-1}=\lim_{z \to0}\frac{z^3+z}{e^{\frac {z}{10}}-1}=10\lim_{z \to0}\frac{3z^2+1}{e^{\frac {z}{10}}}=10\neq0$$
Where l'Hopital's rule was used in the second equality, so $z=0$ is a simple pole of the integrand. Therefore
$$\text{Res}_{z=0} \frac{z^2+1}{e^{\frac {z}{10}}-1}=\lim_{z \to0}\frac{z^3+z}{e^{\frac {z}{10}}-1}=10$$
as calculated above and so using Residue theorem$$\oint_C \frac{z^2+1}{e^{\frac {z}{10}}-1}=2\pi i \text{Res}_{z=0} \frac{z^2+1}{e^{\frac {z}{10}}-1}=20\pi i$$
A: We can write
$$\begin{align}
\oint_{|z|=1}\frac{z^2+1}{e^{z/10}-1}\,dz&=2\pi i \text{Res}\left(\frac{z^2+1}{e^{z/10}-1},z=0\right)\\\\
&=2\pi i \lim_{z\to 0}\frac{z(z^2+1)}{e^{z/10}-1}\\\\
&=2\pi i \lim_{z\to 0}\frac{(z^2+1)+2z^2}{(1/10)e^{z/10}}\\\\
&=20\pi i
\end{align}$$
A: It has a simple pole at the origin because $(\mathrm e^{z/10}-1)' = \frac{1}{10}\mathrm e^{z/10} \neq 0$ when $z=0$.
You need to find 
$$\lim_{z \to 0} z \times \frac{z^2+1}{\mathrm e^{z/10}-1} = \lim_{z \to 0}\frac{z^3+z}{\mathrm e^{z/10}-1}$$
To do this, use L'Hopital's rule:
$$\lim_{z \to 0}\frac{z^3+z}{\mathrm e^{z/10}-1} = \lim_{z \to 0}\frac{3z^2+1}{\frac{1}{10}\mathrm e^{z/10}}=10$$
