# Сontour integral

everyone. Can you help with the task?

## migrated from mathematica.stackexchange.comJun 12 at 17:53

This question came from our site for users of Wolfram Mathematica.

Mathematica is great at checking contour integration results, even very complicated ones. If you're taking Complex Analysis, it's a good idea to learn how to check your work. This is an easy one because by the Residue Theorem, it's zero. But suppose it was a very complicated one and I solved it analytically and found it to be zero. How could I check my work? I would numerically integrate it as follows:

myz[t_] := 1 + 1/2 Exp[I t];
myf[z_] := 1/(z^2 (z^2 + 4));
NIntegrate[myf[z] D[myz[t], t] /. z -> myz[t], {t, 0,
2 \[Pi]}]

2.17925*10^-17 - 3.14419*10^-17 I