If $\iint_G |f(x+iy)|\,dx \, dy<\infty$ then $f$ has a removable singularity at $z=a$ The following is a question take from Conway's:

Exercise $17$ Let $f$ be analytic in the region $G = ann(a;0,R).$ Show that if $\iint_{G}|f(x+iy)|^{2}\,dx\,dy < \infty$ then $f$ has a removable singularity at $z=a$.  Suppose that $p>0$ and $\iint_{G}|f(x+iy)|^{p}\,dx\,dy<\infty$; what can be said about the nature of the singularity at $z=a$?

With a double integral in this problem, I have no idea on how I can even approach this. Any assistance will be appreciated! Thank you in advance. 
 A: Expand $f$ into a Laurent series $f(z) = \sum_{n = -\infty}^{\infty} a_n z^n$. Then for any $\varepsilon > 0$ and $R' < R$ we have
$$ \int_{\varepsilon < |z| < R'} |f(x + iy)|^2 \, dx dy = \int_{\varepsilon < |z| < R'} f(x + iy) \overline{f(x + iy)} \, dx dy = \\
\int_{\varepsilon < |z| < R'} \left( \sum_{n,m} a_n z^n \overline{a_m} \overline{z}^m\right) \, dx dy = \int_{\varepsilon}^{R'} \int_0^{2\pi} \sum_{n,m} a_n (re^{i\theta})^n \overline{a_m} (re^{-i\theta})^m \, r dr d\theta = \\
\sum_{n,m} a_n \overline{a_m} \int_{\varepsilon}^{R'} \int_0^{2\pi} r^{n + m + 1} e^{i(n - m)\theta} \, dr d\theta = 2 \pi \sum_{n} |a_n|^2 \int_{\varepsilon}^{R'} r^{2n + 1} \, dr.$$
where the manipulations with the sums and the integrals are justified by the fact the Laurent series converges uniformly on $\varepsilon \leq |z| \leq R'$. As we let $\varepsilon \to 0$, the integral converges so the right hand must converge. Note that
$$ \lim_{\varepsilon \to 0} \int_{\varepsilon}^{R'} r^{2n + 1} \, dr = \infty $$
for $n < 0$ so Fatuo's lemma implies that $a_n = 0$ for all $n < 0$ and the singularity is removable.

If $p > 2$ and $\int_{A(0,r)} |f(x+iy)|^p \, dx dy < \infty$ then Holder's inequality implies that also $\int_{A(0,r)} |f(x+iy)|^2 \, dx dy < \infty$ so the above also applies. For the case $0 < p < 2$, consider $f(z) = 1/z$ on $A(0,1)$. We have
$$ \int_{A(0,1)} |f(z)|^p \, dx dy = \int_0^1 \int_0^{2\pi} r^{1 - p} \, dr d\theta = \frac{2\pi}{2 - p} < \infty $$
so you can't deduce that the singularity is removable in the case $0 < p < 2$.
