The conversation between human and AI assistant.
How do I show that [math]|\alpha| \leq max[/math] {[math]1, \sum_{i=0}^{n-1} |a_i|[/math]} where [math]\alpha [/math] is zero of [math]p(x)=\sum_{i=0}^{n} a_i x^i,[/math] where [math]a_i[/math] may be complex and [math]a_n =1[/math]?
Yes, that is correct! Can you explain how we can prove the equation?
Sure! First, we know that [math]p(\alpha) = 0[/math] since [math]\alpha[/math] is a zero of [math]p(x)[/math]. Then, we can write [math]|\alpha|[/math] as [math]\frac{|p(\alpha)|}{|a_n|}[/math]. Since [math]a_n = 1[/math], we have [math]|\alpha| = |p(\alpha)|[/math]. Next, we can use the triangle inequality to obtain [math]|p(\alpha)| \leq \sum_{i=0}^{n-1} |a_i||\alpha|^i[/math]. Finally, we can substitute [math]|\alpha|[/math] with [math]|p(\alpha)|[/math] in the inequality to get [math]|p(\alpha)|\leq \max \{1, \sum_{i=0}^{n-1}|a_i|\}[/math], which completes the proof.