This is a question on one of my homework sheets for a complex function theory module on my maths MSc.
I believe I need to use the Laurent series for $f(z)$ around 3 here, so I wrote $f(z)$ as:
$$f\left(z\right)\ =\ \sum_{n=-\infty}^{\infty}a_n\left(z-3\right)^n$$
And then I used the integral representation of the coefficients:
$$a_n=\frac{1}{2\pi i}\int_{S_p^+\left(3\right)}^{ }\frac{f\left(z\right)}{\left(z-3\right)^{n+1}}dz$$
So this would mean that we have:
$$f\left(z\right)\ =\ \left|\sum_{n=-\infty}^{\infty}\left(\frac{1}{2\pi i}\int_{S_p^+\left(3\right)}^{ }\frac{f\left(z\right)}{\left(z-3\right)^{n+1}}dz\right)z^n\right|\le\frac{\left|\left(z-2\right)^2\right|}{\left|z-3\right|}$$
I somehow need to determine how many coefficients of the Laurent series are nonzero, but I'm kind of stuck on where to go next. I noticed that by Cauchy's nth Derivative Formula that, since
$$f^n\left(3\right)=\frac{n!}{2\pi i}\int_{S_p^+\left(3\right)}^{ }\frac{f\left(z\right)}{\left(z-3\right)^{n+1}}dz$$
I guess that means we could write $a_n=\frac{f^n\left(3\right)}{n!}$, but I don't know if that's helpful here.
Clearly, $$ g(z)=(z-3)f(z), $$ has a removable singularity at $z=3$, and hence it is extended to an entire function. Call its extension $g$ as well.
Then $$ |g(z)|\le |z-2|^2, \quad \text{for all $z\in\mathbb C$}. $$ If we set $$ h(z)=\frac{g(z)}{(z-2)^2}, \quad z\ne 2, $$ then $|h(z)|\le 1$, for all $z\ne 2$, and therefore $h$ has a removable singularity at $z=2$. Hence $h$ is extended to an entire function, which is bounded by $M=1$. Therefore, $h$ is constant, $h(z)\equiv c$, with $|c|\le 1$.
Consequently $$ f(z)=\frac{c(z-2)^2}{z-3}, \quad z\in\mathbb C\setminus\{3\}. $$ for some $c$, with $|c|\le 1$.