Consider $ z^d + a_{d-1} z^{d-1} + \ldots + a_1 z + a_0 $ with each $ a_j \in \mathbb{C} $. Any root $ c \in \mathbb{C} $ must satisfy $ | c | \leq 1 + | a_{d-1} | + \ldots + | a_0 | $ (i.e. must be in the closed disk $ | z | \leq 1 + \sum | a_j | $)

Proof is easy. Roots with $ | c | < 1 $ will obviously satisfy.
For those with $ | c | \geq 1 $ we proceed as :
$ c^d = -(a_{d-1} c^{d-1} + \ldots + a_0) $, so $ | c | = \left| a_{d-1} + \dfrac{a_{d-2}}{c} + \ldots + \dfrac{a_0}{c^{d-1}} \right| $ $ \leq \left| a_{d-1} \right| + \left| \dfrac{a_{d-2}}{c} \right| + \ldots + \left| \dfrac{a_0}{c^{d-1}} \right| $ $ \leq | a_{d-1} | + \ldots + | a_0 | .$