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I know that the function $ e^{\frac{1}{z}}$ has an essential singularity in the origin, but I want to study the function

$$ f(z) = \frac{e^{\frac{1}{z}}}{z^2} $$

Especially using the corollary that states

$$ \sum_{j=1}^n Res(f(z);z_j) + Res(f(z);\infty)=0$$

Where $\sum_{j=1}^n Res(f(z);z_j)$ is the sum of all the residues in the finite.

Kowing that setting $w=1/z$ and $g(w) = f(w)/w^2$ we can demonstrate that

$$ Res(f(z);\infty) = -Res(g(w);w=0) $$

Using this on the function I want to study we have

$$ g(w)=e^w $$

That doesn't have singularity at $w=0$, so we can say that $Res(f(z);\infty) = 0$

Which means that the original function too doesn't have a residue at $z=0$.

I tried to see if even in the Laurent expansion this is true, and we can write, knowing that $ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $:

$$ f(x) = \sum_{2}^{\infty} \frac{1}{n!\cdot x^n} $$

Which doesn't have the term for $x^{-1}$, therefore there's no residue.

But it doesn't make sense to me, because even from the Laurent expansion we can see that it still has an essential singularity in $z=0$.

I either got some calculations wrong or I didn't get something right.

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  • $\begingroup$ Origin is an essential singularity with residue $0$. Why is that a contradiction? $\endgroup$ Commented Jun 11 at 11:27
  • $\begingroup$ @geetha290krm so a function can have residue 0 at a singularity? Does that mean that no matter on which contour I integrate the function around the origin I have $ \oint_{\gamma} f(z)\;dz = 0$? $\endgroup$ Commented Jun 11 at 11:54
  • $\begingroup$ Yes, for sure. $e^{1/z^{2}}$ is another such example. $\endgroup$ Commented Jun 11 at 12:01
  • $\begingroup$ There is nothing wrong for an isolated singularity (even essential ones) to have residue $0$. Plus, if we replace $z$ by $\frac{1}{z}$ in $\frac{e^{1/z}}{z^2}$, we get $z^2e^z$ which is an entire function, which doesn't look as pathological, but it's really the same function (geometrically), with $0$ and $\infty$ switched. $\endgroup$ Commented Jun 13 at 7:43

2 Answers 2

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When $0<|z|<\infty$, $f(z)$ has Laurent expansion: $$f(z)=\frac{e^{\frac{1}{z}}}{z^2} =\frac{1}{z^2}\sum_{n=0}^{\infty}\frac{1}{n!}\frac{1}{z^n} =\frac{1}{z^2}+\frac{1}{z^3}+\cdots.$$ So $$Res(f(z);\infty)=-a_{-1}=0,\quad Res(f(z);0)=a_{-1}=0,$$ where $a_{-1}$ is the coefficient of $\frac1z$ in the Laurent expansion. $\infty$ is removeable singularity of $f$ and $0$ is essential singularity of $f$.

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By the result that, for $r > 0$ $$ \int_{\partial D_r(0)} z^n = \begin{cases} 2\pi i & n = -1\\ 0 & \text{else} \end{cases} $$ one can see that, given a Laurent expansion, integrating is just ''reading off'' the $a_{-1}$ term of the Laurent expansion. This doesn't reflect anything about the nature of the singularity, as there's examples of functions with poles: $$ \int_{\partial \mathbb{D}} \frac{1}{z^2} = 0 \quad \text{ with } 0 \text{ order 2 pole} $$ and with essential singularities: $$ \int_{\partial \mathbb{D}} e^{1/z^2} = 0 \text{ with } 0 \text{ essential singularity} $$ (as pointed out by @geetha290krm) which have integrals that vanish.

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