Singularities of complex function $\frac{e^{\frac{1}{z}}}{z^2}$

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.

• Origin is an essential singularity with residue $0$. Why is that a contradiction? Commented Jun 11 at 11:27
• @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 $\oint_{\gamma} f(z)\;dz = 0$? Commented Jun 11 at 11:54
• Yes, for sure. $e^{1/z^{2}}$ is another such example. Commented Jun 11 at 12:01
• 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. Commented Jun 13 at 7:43

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 $$f$$.
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.