Suppose that a function $f$ is analytic and bounded in $B(z_0;R)-\{z_0\}$ of a point $z_0$. Show that the Laurent series expansion of $f(z)$ in the region $0 < |z-z_0| < r < R$ does not involve negative power of $(z-z_0)$. This means $z_0$ is a removable singularity of $f(z)$.
$f$ is analytic and bounded in $B(z_0,R)-\{z_0\}$
In the region $0< |z-z_0| <R$, Laurent expansion is allowed since $f(z)$ is indeed analytic in the annulus. Hence the Laurent series is $$f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n + \sum_{n=1}^\infty b_n(z-z_0)^{-n}$$
Our aim is to show that $b_n = 0$ for all $n$ and for all $r < R$.
Since $f$ is bounded, we have $|f(z)| < M$.
Then it suffices to show $|b_n| = 0$.
Indeed by formula: $$|b_n| = \left|\dfrac{1}{2\pi i}\int_C(z_0,R)\dfrac{f(z)}{(z-z_0)^{-n+1}}dz\right|$$
Now on $|z-z_0| = r$, we have $$\left|\dfrac{f(z)}{(z-z_0)^{-n+1}}\right| \leq \dfrac{M}{r^{-n+1}}$$
The above follows because $|f(z)| \leq M$ and $|(z-z_0)^{-n+1}| = r^{-n+1}$ (Recall that $|z-z_0|=r$.
Now we apply the ML formula, one shoud observe the length of the $C$ of radius $r$ is just $2\pi r$. Hence $$|b_n| = \left|\dfrac{1}{2\pi i}\int_C(z_0,R)\dfrac{f(z)}{(z-z_0)^{-n+1}}dz\right| \leq \left|\dfrac{1}{2\pi i}\right| 2\pi r \cdot \dfrac{M}{-r^{n+1}} = Mr^n$$
And as $r \rightarrow 0, Mr^n \rightarrow 0$ and hence $|b_n| = 0 \implies b_n = 0 ~~\forall n$ and thus the Laurent series of $f(z)$ at $z_0$ has no negative powers of $(z-z_0)$.
Ok so here is my answer and it is correct because i mimic previous questions i did. One thing i do not understand till now is why can we assume the $r$ tend to $0$ part. It will not be true if $r$ does not tend to $0$, won't it? So i feel the answer is a bit counter intuitive.
