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| visits | member for | 1 year, 1 month |
| seen | Jan 21 at 12:17 | |
| stats | profile views | 103 |
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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) |
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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? |
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Jul 24 |
revised |
Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ added 29 characters in body |
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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. |
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Jul 24 |
revised |
Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ edited title |
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Jul 24 |
asked | Laurent series - $f(z)=\frac{1}{z^2-4}+\frac{1}{6-z}$ |
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Jul 23 |
asked | Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ |
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Jul 23 |
comment |
Complex contour integral "it cannot apply to this case as we only have one quarter of a circumference" - because $t \in \left[0,\frac{\pi}{2}\right]$? |
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Jul 23 |
comment |
Complex contour integral Could you elaborate your question, I am not sure in which direction you are pointing... |
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Jul 23 |
asked | Complex contour integral |
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Jul 23 |
asked | Evaluating $\int_{|z|=1} \sin\left(e^{\frac{1}{z}}\right) \ dz$ |
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Jul 23 |
comment |
Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ Thank you very much! |
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Jul 23 |
accepted | Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ |
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Jul 23 |
comment |
Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ The residue theorem? |
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Jul 23 |
asked | Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ |
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Jul 23 |
comment |
Evaluating $\int_\gamma \frac{\cos(z)}{z^3+2z^2} \ dz$ So we have $f'(0)=-\frac{1}{4}$ and $\int_\gamma \frac{\cos(z)}{z^3+2z^2} dz=-\frac{1}{2}\pi i$. Thank you very much! |
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Jul 23 |
revised |
Evaluating $\int_\gamma \frac{\cos(z)}{z^3+2z^2} \ dz$ added 2 characters in body |
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Jul 23 |
asked | Evaluating $\int_\gamma \frac{\cos(z)}{z^3+2z^2} \ dz$ |
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Jul 23 |
accepted | Evaluating $ \int_{|z|=1} \frac{e^z}{(z+3)\sin(2z)} \ dz$ |
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Jul 23 |
comment |
Evaluating $ \int_{|z|=1} \frac{e^z}{(z+3)\sin(2z)} \ dz$ It's a removable singularity. So we could look at $\lim_{z\rightarrow 0} f(z) = \frac{1}{6}$. Do we now set $f(0)=\frac{1}{6}$ and carry on with the Cauchy integral formula? What would have happened if the singularity wasn't removable or if we had multiple singularities? |
