Let C be the positively oriented circle in the complex plane of radius 3 centered at the origin. What is the value of the integral

Correct option is D

Given:

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• The integral is over a positively oriented circle CCC with radius 3 centered at the origin: 

  

​Solution: 

$$e^z - e^{-z} = 0 \implies e^{2z} = 1$$

$$e^{2z} = e^{i(2k\pi)}, \, k \in \mathbb{I} \implies 2z = 2k\pi i$$

$$z = k\pi i; \, k \in \mathbb{I}$$

$$z = 0, \pm \pi i, \pm 2\pi i, \dots$$

$$\text{Inside } |z - 0| = 3, \text{ we have only one singular point } 0 \text{ of } \frac{1}{z^2 (e^z - e^{-z})}, \text{ which is also a pole.}$$

$$\text{of order 3, so } \text{Res}_{z=0} \frac{1}{z^2}\left(e^z - e^{-z}\right) = \\\ \\= \text{Res}_{z=0} \frac{1}{z^2} \left[ \left(1 + z + \frac{z^2}{2} + \dots \right) - \left(1 - z + \frac{z^2}{2} - \dots \right) \right] \\\ \\= \text{Res}_{z=0} \frac{1}{z^2} \left( 2z + \frac{z^3}{3} + \frac{z^5}{60} + \dots \right)\\\ \\= \text{Res}_{z=0} \frac{1}{2z^3} \left( 1 + \frac{z^2}{6} + \frac{z^4}{120} + \dots \right)^{-1}\\\ \\= \text{Res}_{z=0} \frac{1}{2z^3} \left( 1 - \frac{z^2}{6} - \frac{z^4}{120} + \dots + \left( \frac{z^2}{6} + \frac{z^4}{120} + \dots \right)^2 + \dots \right)= \text{Res}_{z=0} \frac{-1}{12z}\\\ \\= -\frac{1}{12} = \text{coefficient of } \frac{1}{z}\\\ \\\text{So, } \int_C \frac{dz}{z^2 \left(e^z - e^{-z}\right)} = 2\pi i \left(-\frac{1}{12}\right)\\\ \\ = -\frac{\pi i}{6}$$​​​​​​​