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Let $f(z)$ be analytic for $|z|>r$ and let it be bounded $|f(z)|\leq M, M>0$ wherever it is analytic. Show that the coefficients of the Laurent Series of $f(z)$ are $0$ for $j\geq 1$.

I have found two approaches to solve this. I'm not sure about this one:

The positive part of the Laurent Series:

$$\sum_{j=0}^\infty a_j(z-z_0)^j$$

converges when $|z-z_0|<R$. In our case $R$ is infinity so the only way for the sum to converge is if $a_j=0$. Does this make sense?

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The function $F(z) = f(1/z)$ is holomorphic for $0 < |z| < 1/r$, where it is uniformly bounded by $M$. So $F$ has a removable singularity at $0$, which gives a power series expansion $$ F(z) = \sum_{n=0}^{\infty}b_nz^n,\;\;\; 0 < |z| < 1/r. $$ Therefore, $$ f(z) = F(1/z) = \sum_{n=0}^{\infty}b_n z^{-n},\;\;\; |z| > r. $$ Your argument doesn't work because you're not addressing boundedness. A series can convergence everywhere in $\mathbb{C}$.

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