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I'm trying to solve an exercise about the classification of singolarity of a function.

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

I have found two poles. A first order pole in $i$ and a second order pole in $-i$. There should also be an essential singularity. How can I calculate it? Thank you so much.

Enzos Gint
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  • Why do you think there's an essential singularity? Unless you count the infinity point there's no more singularities. – skyking May 19 '17 at 10:31
  • yes, i have to evaluate it too infinity. But if i change $z$ with $w^-1$ and considering the $lim_{z\to0}{\frac{e^{\frac{i}{z}+1}-1}{(\frac{1}{z^2}+1)^2}} $the limit exisist and therefore there is not essential singularity – Enzos Gint May 19 '17 at 10:45
  • That's an interresting finding, how do you come to the conclusion that the limit exists? I would say that it doesn't exist (because $e^z$ take both superpolynimially large and small values near infinity). – skyking May 19 '17 at 11:02

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The function has a essential singularity at infinity, you can see this by noting that the poles can be eliminated by subtraction with rational functions that is:

$$\phi(z) = f(z) - q(z)$$

have no finite singularities, consequently it can be Taylor expanded with infinite convergence radius. That is:

$$\phi(z) = \sum c_kz^k$$

Now either the sum is finite (in which case we have a pole at infinity or it's constant) or is infinite (in which case we have a essential singularity at infinity).

Suppose it were finite then we would have that $\phi(z)$ were a polynomial and that

$$e^{iz+1} = (z+1)^2 f(z) + 1 = (z+1)^2 (\phi(z)+q(z)) + 1$$

that is $e^z$ would be a rational function and since $e^z$ have no poles it would be a polynomial which we know it isn't. So we can conclude the Taylor expansion to be infinite and we have a essential singularity at infinity.

skyking
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