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Classify all the singularities of $f(z) = \frac{1}{(z-2)^2}+e^{\frac{1}{3-z}}$

I know that the singularities of $f$ are exactly $z=2,z=3$. I just want to check that if my solution for the classification is correct:

$z=2$: since $1/(3-z)$ is analytic at $z=2$, then the composition $e^{\frac{1}{3-z}}$ is also analytic at this point. Analyitic functions admit Taylor expansion, hence the expansion of $e^{\frac{1}{3-z}}$ will only give "positive coeffcients" (coefficients with positive powers), so it suffices to only look for the Laurent expansion of $\frac{1}{(z-2)^2}$, which is already in this form. Thus, $z=2$ is a pole of order 2.

$z=3$: now $\frac{1}{(z-2)^2}$ is analytic at this point, so by the argument above we don't have to care about its series expansion. Now, since $e^z$ is entire, this series admits expansion in the region $3<|z|<\infty$, so we can write:

$$e^{\frac{1}{3-z}} = \sum_{n=0}^{\infty}\frac{1}{n!(3-z)^n}.$$

The uniqueness of the laurent series guarantee us that this series above is the Laurent expansion of $e^{\frac{1}{3-z}}.$ Therefore, $z=3$ is an essential singularity.

Is it correct? Or I do have to find the series expansion in each case?

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I am with you solidly until you claim to only be concerned about the principal (negative powers of $z$) terms in the Laurent series of $f$. I think you're using some facts about sums of Laurent series that you're not stating. It is not guaranteed, in general, that the sum of two Laurent series will converge.
I'm not sure how to fix this problem because in my class we have not done this type of thing either, or maybe in this case it happens to converge and it's not a problem luckily(probably true since these are nice functions).

But as an alternative, I think you can just rewrite the function as $$ f(z) = \frac{1+e^{\frac{1}{3-z}}(z-2)^2}{(z-2)^2} = (1+e^{\frac{1}{3-z}}(z-2)^2)\frac{1}{(z-2)^2} $$ The general theory states that if we can factor $f$ into the form $$ f(z) = \frac{\phi(z)}{(z-z_0)^m} $$ then $z_0$ is a pole of order $m$, if $\phi$ is analytic and non-zero at $z_0$. We see $f$ is conveniently already in this form, and $\phi(z)$ is analytic and non-zero at $2$, thus $z =2$ is a pole of order 2.

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    $\begingroup$ And an essential singular point at $z = 3$. $\endgroup$ – David Nov 23 '18 at 2:40
  • $\begingroup$ Ok, that's a good "trick" to definetly prove that $z=2$ is a pole, thanks. And what about my proof of $z=3$ being an essential singularity? bascially I am using the same argument again $\endgroup$ – user2345678 Nov 23 '18 at 14:26
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Your reasoning is correct.

$e^{-\frac{1}{z-3}}$ is analytic at $z=2$, so it can be expanded in a Taylor series about that point in the region $0<|z-2|<1$:

\begin{align}e^{-\frac{1}{z-3}}=e+e(z-2)+\frac{3e}{2}(z-2)^2+\cdots\end{align}

Therefore, the Laurent series of $f(z)$ about $z=2$ in the region $0<|z-2|<1$ is given by:

\begin{align}f(z)=&\frac{1}{(z-2)^2}+e+e(z-2)+\frac{3e}{2}(z-2)^2\\&+\cdots\end{align}

Similarly, $\frac{1}{(z-2)^2}$ is analytic at $z=3$, so it can be expanded in a Taylor series about that point in the region $0<|z-3|<1$:

\begin{align}\frac{1}{(z-2)^2}=&1-2(z-3)+3(z-3)^2+\cdots\end{align}

Therefore, the Laurent series of $f(z)$ about $z=3$ in the region $0<|z-3|<1$ is given by:

\begin{align}f(z)=&\cdots+\frac{1}{2}\frac{1}{(z-3)^2}-\frac{1}{z-3}+2\\&-2(z-3)+3(z-3)^2+\cdots\,\end{align}

and we arrive at your conclusions.

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