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I would like some help integrating the complex function $$\frac{\tan(z) + 1}{z^2-z}$$ along the contour $|z| = 1/2$.

I factored the denominator and then thought of setting $(\tan(z) + 1)/z$ as $f(z)$ but then it is not analytic inside the contour (i.e. not analytic at z = 0) so I cannot use the Cauchy integral formula. Any ideas?

saulspatz
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Elli
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1 Answers1

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Let $\gamma$ be the contour $|z|=\frac12$. $$\begin{align} \int_\gamma\frac{\tan z+ 1 }{z^2-z}\,\mathrm{d} z&=\int_\gamma\frac{\tan z}{z(z-1)}\,\mathrm{d}z+\int_\gamma\frac1{z(z-1)}\,\mathrm{d}z\\&=0+\int_\gamma\frac{\mathrm{d}z}{z-1}-\int_\gamma\frac{\mathrm{d}z}{z} \end{align}$$

The first integral becomes $0$ because the integrand is analytic on and within $\gamma$, since $\frac{\tan z}z$ has a removable singularity at $0$. I'm sure you can finish it from here.

saulspatz
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  • I thought that since $(tanz)/z$ has a removable singularity at 0, then the integrand is not analytic within $γ$? To finish it off, we would use the cauchy-integral formula on $dz/(z-1)$ right? – Elli May 30 '21 at 23:34
  • @Elli The removable singularity doesn't affect the integral. By definition, the integral depends only on the function value on the path of integration. Removing the singularity doesn't affect those. $1/(z-1)$ is analytic on and within $\gamma$. You need to use the Cauchy integral formula on $\int_\gamma\frac{\mathrm{d}z}{z}$ – saulspatz May 31 '21 at 04:19