Using Cauchy's integral theorem without polynomial in denominator

Question

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.

Answer 1

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$$\oint_C \frac{z^2+1}{e^{\frac {z}{10}}-1}=2\pi i \text{Res}_{z=0} \frac{z^2+1}{e^{\frac {z}{10}}-1}=20\pi i$$

Answer 2

We can write

$$\begin{align} \oint_{|z|=1}\frac{z^2+1}{e^{z/10}-1}\,dz&=2\pi i \text{Res}\left(\frac{z^2+1}{e^{z/10}-1},z=0\right)\\\\ &=2\pi i \lim_{z\to 0}\frac{z(z^2+1)}{e^{z/10}-1}\\\\ &=2\pi i \lim_{z\to 0}\frac{(z^2+1)+2z^2}{(1/10)e^{z/10}}\\\\ &=20\pi i \end{align}$$

Answer 3

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: $$\lim_{z \to 0}\frac{z^3+z}{\mathrm e^{z/10}-1} = \lim_{z \to 0}\frac{3z^2+1}{\frac{1}{10}\mathrm e^{z/10}}=10$$