Chris
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 Jul 24 accepted Evaluating $\int_{-\pi}^{\pi} \cos(e^{it})dt$ Jul 24 accepted Complex contour integral Jul 24 comment Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Thank you very much! Jul 24 comment Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Alright, thank you! Jul 24 revised Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ added 4 characters in body Jul 24 accepted Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Jul 24 comment Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ Great, things are much clearer now. I was until now utterly confused about that! Jul 24 accepted Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ Jul 24 comment Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ That's very good robjohn, thank you so much! Question on point 3: In my case, there are no negative powers of $z-0$, doesn't this fact change something? Do we still have a Laurent series, even if all powers are positive ($k$ starts at $0$ in both cases) Jul 24 comment Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ I've added the missing information. So we don't do anything with $\frac{1}{z-2}$, because it's already part of the expansion around z=2 and there is nothing additional to be done here, right? Jul 24 revised Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ added 29 characters in body Jul 24 comment Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Some exercise notes from a complex analysis course I attended. Jul 24 revised Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ edited title Jul 24 asked Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Jul 23 asked Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ Jul 23 comment Complex contour integral "it cannot apply to this case as we only have one quarter of a circumference" - because $t \in \left[0,\frac{\pi}{2}\right]$? Jul 23 comment Complex contour integral Could you elaborate your question, I am not sure in which direction you are pointing... Jul 23 asked Complex contour integral Jul 23 asked Evaluating $\int_{|z|=1} \sin\left(e^{\frac{1}{z}}\right) \ dz$ Jul 23 comment Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ Thank you very much!