$$ f(z)=\frac{z^3}{e^z-1} $$
- Is this a simple pole at $z=0$ or some other types of pole?
- If it is a simple pole, what is its residue?
- Is it using this formula or other else? $$ \lim_{z\to 0}=zf(z) $$
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K. Rmth
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user111605
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What is the order of the zero of $e^z-1$ in $0$? – Daniel Fischer Nov 26 '13 at 19:30
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so it is a simple pole at z=0? – user111605 Nov 26 '13 at 19:37
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Not so fast. How does the numerator, $z^3$ behave in $0$? – Daniel Fischer Nov 26 '13 at 19:39
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both z^3 and e^z-1 be zero when z=0 – user111605 Nov 26 '13 at 19:49
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And what happens then? How do you determine whether it's a pole or a removable singularity? – Daniel Fischer Nov 26 '13 at 19:50
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It is not removable but I am not sure is this a simple pole or essential... – user111605 Nov 26 '13 at 19:54
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Why do you think it's not removable? – Daniel Fischer Nov 26 '13 at 19:55
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It seems that I have messed up my concept.. why it is removable? – user111605 Nov 26 '13 at 19:57
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1Write $e^z - 1 = (z-0)\cdot g(z)$, where $g$ is holomorphic with $g(0) \neq 0$. Then $$f(z) = \frac{z^3}{e^z-1} = \frac{z^3}{z\cdot g(z)} = \frac{z^2}{g(z)},$$ and that shows that $f$ not only has a removable singularity in $0$, it even has a zero of order $2$ (which is $3 - 1$) in $0$. – Daniel Fischer Nov 26 '13 at 20:03
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$$e^z=\sum_{k=0}^\infty\frac{x^k}{k!}\implies\frac{z^3}{e^z-1}=\frac{z^3}{z+\frac{z^2}2+\frac{z^3}6+\mathcal O(z^4)}=$$
$$=\frac{z^2}{1+\frac z2+\frac{z^2}6+\mathcal O(z^3)}$$
The above shows at once that $\;z=0\;$ is a removable singularity...though the last step requires a little justification (factor out $\;z\;$, and then you can cancel. Why?)
DonAntonio
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