Chris
Reputation
561
Next privilege 1,000 Rep.
Create tags
 Jul24 comment Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Thank you very much! Jul24 comment Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Alright, thank you! Jul24 revised Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ added 4 characters in body Jul24 accepted Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Jul24 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! Jul24 accepted Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ Jul24 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) Jul24 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? Jul24 revised Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ added 29 characters in body Jul24 comment Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Some exercise notes from a complex analysis course I attended. Jul24 revised Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ edited title Jul24 asked Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ Jul23 asked Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ Jul23 comment Complex contour integral "it cannot apply to this case as we only have one quarter of a circumference" - because $t \in \left[0,\frac{\pi}{2}\right]$? Jul23 comment Complex contour integral Could you elaborate your question, I am not sure in which direction you are pointing... Jul23 asked Complex contour integral Jul23 asked Evaluating $\int_{|z|=1} \sin\left(e^{\frac{1}{z}}\right) \ dz$ Jul23 comment Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ Thank you very much! Jul23 accepted Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ Jul23 comment Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ The residue theorem?