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Residue theorem and Cauchy's Integral Formula for $e^{\frac {-3} {z^2}} $

I'm looking at an exam question and one of the questions asks

find the residue of $f(z)=e^{\frac {-3} {z^2}} $ at $z=0$ which is equal to 0

however further on in the question it then asks to find the residue of

Evaluate $\int_Ce^{\frac {-3} {z^2}} dz$ where C is the unit circle.

Is this not simply just equal to 0 also? Thanks

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    Indeed, the residue and thus the integral are both zero. This is coincidental, no theorems like Cauchy-Goursat are causing this. – 2017-01-06
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    See http://math.stackexchange.com/questions/2086047/finding-the-residue-of-e-frac-3-z2?rq=1 – 2017-01-07

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Well how did you define the residue? It s the first "negative" coefficient of the Laurent series expansion. You just have to use here the series expansion of the exponential function! =)