Pole and residue of the following function at infinity

I am confused about one particular problem regarding complex infinities. Suppose i have EXP[-z^2] with z being the complex number. Clearly it has poles at z=+i(infinity) and -i(infinity). . How to show that. If I replace w=1/Z then surely it doesn't work. So what should I do. Also I have a general question. When we talk about poles do we view all things in on Riemann surface .Then i(infinity) and-i (infinity) both correspond to the same pont (north pole). I am sorry for asking such sort of questions. but I am really not getting the point.

• I think $\exp(-z^2)$ has an essential singularity at $\infty$. – GEdgar Sep 16 '15 at 15:14
• @ GEdgar Can you tell me how can you see that? – kau Sep 16 '15 at 16:08
• Same reason that $e^z$ has an essential singularity at $\infty$. At a pole, $|f(z)| \to \infty$ in all directions of approach to the point. But here, $|f(z)| \to 0$ in some directions and $|f(z)| \to \infty$ in other directions. – GEdgar Sep 16 '15 at 17:48
• @ GEdgar ...Yeah that's right..But then let me ask you this question... So whenever we say something about infinity we always think of them to one single point,namely the north pole of the riemann sphere. So I do not need to worry which infinity it is. And I can safely say that it is non analytic at infinity (doesn't matter which direction,since all of them are same on riemann sphere.).. Is this right??? If it is then We generally discuss about the poles taking some w=1/z and then see the behavior at w->0.. I can't do that here. So what would be the actual rigorous way to show it in this case? – kau Sep 17 '15 at 18:35
• Same method. Let $w=1/z$. Then $\exp(-1/w^2)$ has an essential singularity at $w=0$. – GEdgar Sep 17 '15 at 20:46