How i can show that if $\lim_{z\rightarrow\infty}\frac{|f(z)|}{|z|^{m}}=0$ then $f(z)$ is a polynomial of degree $< m$? Note: $f(z)$ is a polynomial in $\mathbb{C}[x].$ Thanks,
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Hint: Write $f(z)$ as $$f(z) = z^{d} \left( a_0 + a_1 z^{-1} + a_2z^{-2} + \dots a_d z^{-d} \right) = z^{d} g(z),$$ where $d = \operatorname{deg} f$ so that $a_0 \neq 0$. Then $g(z) \to a_0$ as $z \to \infty$. Consider $\frac{f(z)}{z^m} = \frac{1}{z^{m-d}} g(z) \sim \frac{a_0}{z^{m-d}}$. This will converge to $0$ iff $m > d$.

Jakub Konieczny
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You cannot assume that $f$ is a polynomial. – lhf Nov 23 '13 at 16:05