Please check Cauchy Integral Formula excercise

[ognik]

Find $ \oint\frac{e^{iz}}{z^3}dz $ where contour is a square, center 0, sides > 1

There is an interior pole of order 3 at z=0

CIF: $ \oint\frac{f(z)}{(z-z_0)^{n+1}}dz = \frac{2\pi i}{n!} f^{(n)}(z_0) = \frac{2 \pi i}{2}f''(z_0) = -\pi i $

 

[alyafey22]

You can check your solution by using the Laurent expansion and looking for the coefficient of $z^{-1}$.