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Determine and find residues for all singular points $z\in \mathbb{C}$ for

(i) $\frac{1}{z\sin(2z)}$

(ii) $\frac{1}{1-e^{-z}}$

Note: I have worked out (i), but (ii) seems still not easy.

Martin
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user31899
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2 Answers2

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Related problems: (I), (II), (III). Just use the following test, a function $f(z)$ has a pole of order $m \in \mathbb{N} $ if

$$ \lim_{z\to z_0}(z-z_0)^m f(z) = c \neq 0, $$

where c is a finite number. For instance, in your case, $z=0$ is a pole of order $2$, since

$$ \lim_{z\to z_0}z^2 \frac{1}{z\sin(2z)} = \frac{1}{2}.$$

For finding residues see here.

Community
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Mhenni Benghorbal
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Thank Mhenni Benghorbal for the hint! I have worked out (i) so far:

The singular points for (i) are $\frac{k\pi}{2}, k \in \mathbb{Z}$,
The case $k=0$ was justified by Mhenni, For $k \neq 0$ the singular points are simple poles, since $$\lim_{z \to \frac{k\pi}{2}}\frac{z-\frac{k\pi}{2}}{z\sin(2z)}=\lim_{z \to \frac{k\pi}{2}}(-1)^k\frac{z-\frac{k\pi}{2}}{z\sin(2(z-\frac{k\pi}{2}))}=(-1)^k\frac{1}{k\pi}\neq0$$ which also gives the corresponding residues.

For (ii): CLearly the singularities are $2ki\pi, k\in\mathbb{Z}$, which are isolated poles (as their reciprocal corresponding to isolated zeros). But I have problem about computing the limit such as $$\lim_{z \to 0} \frac{z e^z}{e^z-1}.$$
Note that L-Hospitial's rule may not apply here, the method by series expansion is not applicable here either.?

Gary
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user31899
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    $$\lim_{z \to 0} \frac{z e^z}{e^z-1}$$ $ \text{It is }, \frac 0 0$ form and so L'hospital's rule is applicable. Now,$$\lim_{z \to 0} \frac{z e^z}{e^z-1}=\lim_{z \to 0} \frac{\frac d {dz} (ze^z)}{\frac d {dz} (e^z-1)}=\lim_{z \to 0} \frac{e^z(1+z)}{e^z}=\lim_{z \to 0}(1+z)=1.$$ – learner Aug 27 '13 at 18:46