Please check this complex integral (#2)

[ognik]

An old exam question is: Evaluate $ \oint \frac{e^{iz}}{z^3}dz $ where the contour is a square of sides a, centered at 0. This has a simple pole of order 3 at z = 0

Perhaps using residues, $ Res(f,0) = \frac{1}{2!}\lim_{{z}\to{0}}\d{^2{}}{{z}^2}z^2 \frac{e^{iz}}{z^3} = \frac{1}{2}\lim_{{z}\to{0}} \d{^2{}}{{z}^2}\frac{e^{iz}}{z} $
But this will have z to some power in the denominator, with z -> 0, so I don't think I can do that

Instead I'll try the Cauchy Integral formula, $ \oint \frac{e^{iz}}{z^3}dz = \frac{2 \pi i}{2!}f^{(3)}(0) $, with $f(z) = e^{iz}$
$f(z) = e^{iz}, f^{(3)}(z)=-ie^{iz}, f^{(3)}(0) = -i$ and $ \oint \frac{e^{iz}}{z^3}dz = \pi $
Is all this correct please?

 

[ThePerfectHacker]

This is wrong. You took an extra derivative.

You can also write $\frac{e^{iz}}{z^3}$ as a power series and notice the only term with $z^{-1}$ in it has coefficient equal to $\frac{(iz)^2}{(2!)z^3} = -\frac{1}{2z}$

When you integrate $z^{-1}$ over that region you get $2\pi i$. Therefore, the final answer will be,
$$ 2\pi i \times -\frac{1}{2} = -\pi i$$

 

[ognik]

All clear, thanks for the series suggestion as well.