Let $F(z)$ be a rational function $\frac{P(z)}{Q(z)}$ such that the degree of $P(z)$ is less than the degree of $Q(z)$ and suppose that all the zeros of $Q(z)$ are contained in the open disk $|z| < r$.
I know that if $f(z)$ is analytic for $|z| > r$ and bounded by M > 0 there, that is, $|f(z)|{\le}M$ for all $z$ with $|z| > r$, then the coefficients of the Laurent series of $f(z)$ for $|z| > r$ satisfy $aj = 0$ for $j = 1,2,3,$....
I'm supposed to show that the coefficients of the Laurent series of $F(z)$ for $|z| > r$ satisfy $aj = 0$ for $j = 1,2,3,$.... by using the corollary above. I know that $F(z)$ is analytic for $|z| > r$ but I'm missing the boundless condition to finish my proof. Any suggestions?
As with any nonconstant polynomial, $|Q(z)|\to\infty$ as $|z|\to\infty.$ Then $\deg(P)<\deg(Q)$ implies $|F(z)|\to 0$ as $|z|\to\infty$ and in particular is bounded for $|z|>r$, say by $M$. The Laurent series of $F$ centered at some $z_0$ has coefficients $$a_k = \frac{1}{2\pi i}\int_{|z-z_0|=R}\frac{F(z)}{(z-z_0)^{k+1}}dz,$$ so $$|a_k|\leq \frac{M}{R^k}$$ for big enough $R$. If $k\geq 1$, you can take $R\to\infty$ to show $a_k = 0.$