AhhH! sorry, of course it's real! .
Anyway, i've now run into a new problem. I can get the integral into the form:
.
What's the best way to tackle this integral?
Yes. Try and get in the habit of checking it with Mathematica if you have access to it:
Code:In[115]:= NIntegrate[(Exp[z^2]/(z - 1)^3)*2*I* Exp[I*t] /. z -> 2*Exp[I*t], {t, 0, 2*Pi}] N[6*E*Pi*I] Out[115]= 1.2434497875801753*^-14 + 51.23840533603834*I Out[116]= 0. + 51.2384053360414*I