MHB Please check Cauchy Integral Formula excercise

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The integral to evaluate is \( \oint\frac{e^{iz}}{z^3}dz \) over a square contour centered at 0 with sides greater than 1. There is a pole of order 3 at \( z=0 \). Applying the Cauchy Integral Formula, the result is calculated as \( -\pi i \) using the second derivative of the function at the pole. Additionally, the solution can be verified by using the Laurent series expansion to find the coefficient of \( z^{-1} \). This confirms the accuracy of the integral evaluation.
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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 $
 
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You can check your solution by using the Laurent expansion and looking for the coefficient of $z^{-1}$.
 

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