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I am trying to evaluate the contour integral $$\oint_{C} \frac{z^{2}}{\sin ^{2} 4 z} d z$$ on a circle $C$ with radius $\pi / 4,$ centred at $z=\frac{1}{4}$ in the complex $z$-plane that is traversed counterclockwise. I have learnt the Cauchy integral formula (for $z_0$ within the contout), $$\frac{1}{2 \pi i} \oint_{C} \frac{f(z)}{z-z_{0}} d z=f\left(z_{0}\right)$$ but don't understand how to approach this integral as most of the ones I have tackled so far were easily reducible to a form that contained $(z-z_0)$ in the denominator. Need some direction to approach such problems.

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Since $\displaystyle\lim_{z\to0}\frac z{\sin(4z)}$ exists (in $\mathbb C$), $0$ is a removable singularity of your function. So, in the disk centered at $\frac14$ with radius $\frac\pi4$, the only singularity that you will have to deal with is $\frac\pi4$. But$$\operatorname{res}_{z=\pi/4}\frac{z^2}{\sin^2(4z)}=\frac\pi{32}$$and therefore, by the residue theorem, your integral is equal to $\frac{\pi^2i}{16}$.

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  • $\begingroup$ Thank you so much. $\endgroup$
    – orionphy
    Commented Jan 16, 2020 at 11:12
  • $\begingroup$ Right! I've edited my answer. Thank you. $\endgroup$ Commented Jan 16, 2020 at 11:14

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