  Let $ x=(x_{n})_{n} $ be a bounded complex sequence and let $M_{x} = \sup_n
|x_n|$. By using a normaloid operator related to the sequence $ x=(x_{n})_{n}
$, we prove that $$ \sup_{\lambda \in \mathbb{C}, |\lambda| \leq M_x} \sup_n
|x_n+\lambda| = 2M_x.$$
