Show $z=0$ is a simple pole or removable singularity if $r\int_{0}^{2\pi}|f(re^{i\theta})|d\theta \lt M$ I am trying to show that if $f$ analytic in $0 \lt |z| \lt R$ and there exists $M \gt 0$ such that for every $0 \lt r \lt R$ we have:
$$r\int_{0}^{2\pi}|f(re^{i\theta})|d\theta \lt M$$
then $z=0$ is either a simple pole or removable singularity.
Since I need to show that $z=0$ is either a simple pole or removable singularity, I tried to prove that $\lim_{z\to0}zf(z) = c$
So I first tried to estimate $|f(z)|$ using Cauchy's integral formula for punctured disk, where $C_1 = r_1e^{i\theta}, C_2 = r_2e^{i\theta}, 0 \lt r_1 \lt r_2 \lt R$:
$$|f(z)| = |\frac{1}{2\pi i} \int_{C_2}\frac{f(\zeta)}{\zeta-z}d\zeta - \frac{1}{2\pi i} \int_{C_1}\frac{f(\zeta)}{\zeta-z}d\zeta | \le$$
$$\le \frac{1}{2\pi}|\int_{C_2}\frac{f(\zeta)}{\zeta-z}d\zeta| + \frac{1}{2\pi} |\int_{C_1}\frac{f(\zeta)}{\zeta-z}d\zeta | =$$
$$\frac{1}{2\pi}|\int_{0}^{2\pi}\frac{r_2ie^{i\theta}f(r_2e^{i\theta})}{r_2e^{i\theta}-z}d\theta| +  \frac{1}{2\pi}|\int_{0}^{2\pi}\frac{r_1ie^{i\theta}f(r_1e^{i\theta})}{r_1e^{i\theta}-z}d\theta| \le$$
$$\le \frac{r_2}{2\pi}\int_{0}^{2\pi}\frac{|f(r_2e^{i\theta})|}{|r_2e^{i\theta}-z|}d\theta +  \frac{r_1}{2\pi}\int_{0}^{2\pi}\frac{|f(r_1e^{i\theta})|}{|r_1e^{i\theta}-z|}d\theta$$
if we choose $r_1 \lt r_2 \lt \frac{1}{2}|z|$ then $|\zeta| \le \frac{|z|}{2}$
and then $|\zeta -z| \ge ||\zeta|-|z|| \ge \frac{|z|}{2} \implies \frac{1}{|\zeta-z|} \le \frac{2}{|z|}$
therefore we have:
$$
|f(z)| \le \frac{r_2}{\pi|z|}\int_{0}^{2\pi}|f(r_2e^{i\theta})|d\theta +  \frac{r_1}{\pi|z|}\int_{0}^{2\pi}|f(r_1e^{i\theta})|d\theta \lt \frac{2M}{\pi |z|}
$$
so $|zf(z)| \lt \frac{2M}{\pi}$ and therefore the limit is finite and could be zero in which case $z=0$ is removable singularity or non-zero in which case it's a simple pole.
Can someone confirm that this proof is correct? I the general approach is fine but there's a mistake please point it out. If I am completely way off please give a hint regarding how to approach this, but without a full solution please

I was able to find a solution using a different approach:
Using Laurent's theorem we have:
$$a_n = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta^{n+1}}d\zeta$$
where $C=re^{i\theta}$
and then using the given property of $f$:
$$
|a_n| = \left| \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta^{n+1}}d\zeta \right| \le \frac{M}{2\pi r^{n+1}}
$$
This expression goes to $0$ as $r \to 0$ for $n<-1$
