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seen Jan 21 '13 at 12:17

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!
Jul
23
accepted Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$
Jul
23
comment Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$
The residue theorem?
Jul
23
asked Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$