# Area integral near essential singularity

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 - en.wikipedia.org/wiki/Casorati%E2%80%93Weierstrass_theorem – Autolatry Apr 7 '12 at 9:00

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