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If we are given a function such as $e^{z+\frac{1}{z}}=e^ze^{\frac{1}{z}},$ and we are trying to classify it's singularities. Is it correct to say, that because $e^{\frac{1}{z}}$ has a singularity at $z_0=0$, and $e^z$ is entire, that f(z) has singularities only at $z_0=0$ ?

Then are we allowed to only consider $e^{\frac{1}{z}}$ when classifying the singularity ?

I.e. during the next step, where we state that because $f(z)=e^{\frac{1}{z}}$ is known to be analytic on all of $\Bbb C /\{0\}$, then by Laurent's theorem we know that because f is differentiable on this set that it can be expressed as

$$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$$

where

$$a_n=\frac{1}{2\pi i}\int_{C(z_0,r)}\frac{f(z)}{(z-z_0)^{n+1}}dz$$

which in the particular case I was asking about would become

$$a_n=\frac{1}{2\pi i } \int_{C(z_0,r)} \frac{e^{\frac{1}{z}}}{z^{n}}dz$$

which we would then parametrise. So is this okay, or should I be considering the entire original function $f(z)=e^{z+\frac{1}{z}}$?

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  • $\begingroup$ I'm lost. What do you mean by "the next step?" You've classified the singularities. What are you trying to do now? $\endgroup$ – saulspatz Apr 22 '18 at 20:36
  • $\begingroup$ @saulspatz All I said was that $z_0=0$, I haven't shown whether it's essential, removable or a pole. That's what I meant by classifying. $\endgroup$ – excalibirr Apr 22 '18 at 20:42
  • $\begingroup$ @saulspatz the next thing I would do after parametrisation would be to find if it exhibits one of three qualities 1) do we have $a_n \neq 0$ when n<0, which would classify it as essential. 2) if the order is some negative value which would imply it's a pole. 3) if its order was positive making it a removable singularity. $\endgroup$ – excalibirr Apr 22 '18 at 20:58
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    $\begingroup$ OK, I understand you now. No, you don't have to worry about $e^z$. Since $e^z\to 1$ at $0$, it's essentially like a constant factor for purposes of determining the type of singularity. $\endgroup$ – saulspatz Apr 22 '18 at 23:04

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