# Evaluating contour integral of a function containing trigonometric complex function in denominator using Cauchy integral formula

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.

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 $$\frac{\pi^2i}{16}$$.

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