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I'm studying for an exam and am stuck on the following. If $f$ is holomorphic on the punctured unit disk $D- \{0\}$, and $0$ is an essential singularity does it follow that

$\displaystyle\int_{D -\{0\}} |f(z)|^{2} dA = \infty$

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While it's good practice to mark edits in your question as such, it seems a bit confusing to say "should say" and then state the same equation again; it makes one wonder whether there are any subtle differences between the two equations. In a case like this where there are no answers or comments yet that refer to the erroneous version, I think it's OK to just correct the question, or perhaps to add "(corrected)" or something like that; I think duplicating the equation does more harm than good. – joriki Apr 7 '12 at 8:54
Agree with @joriki above. not really sure what to do with your question but may I point you to the Casorati - Weierstrass theorem on the behaviour of meromorphic functions near essential singularities - – Autolatry Apr 7 '12 at 9:00
up vote 4 down vote accepted

I believe I found a solution to my own question. I'll write it out to see if anyone agrees, Basically we can write the integral as

$\displaystyle\int_{D -\{0\}} |f(z)|^{2} dA = 2\pi \int_{0}^{1} \sum_{n=-\infty}^{\infty} |a_{n}|^{2}r^{2n+1} dr$

where $a_{n}$ are the Laurent coeficients Here I used Tonelli Thereom along with the fact that $\{e^{i \theta n} | n\in \mathbb{Z} \}$ are orthogonal.

Since we have an essential singularity $a_{k} \ne 0$ for some $k<0$. Thus we have

$\displaystyle\int_{D -\{0\}} |f(z)|^{2} dA = \infty$

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Is it such a formidable effort to write out "coefficients" and "theorem"? I'd estimate that it would take you at least an order of magnitude less time than the sum of all the times that your readers spent decyphering the abbreviations. – joriki Apr 7 '12 at 15:55

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