Why does $| a_n + \frac {a_{n-1}} {z} + \ldots + \frac {a_{0}} {z^n} | \ge | a_n | - |\frac {a_{n-1}} {z} + \ldots + \frac {a_{0}} {z^n} |$ hold?

Question

Why does $$| a_n + \frac {a_{n-1}} {z} + \ldots + \frac {a_{0}} {z^n} | \ge | a_n | - |\frac {a_{n-1}} {z} + \ldots + \frac {a_{0}} {z^n} |$$ hold by the trinagle inequality for $z, a_i \in \mathbb C$ ?

Answer 1

Because $|x+y|\geq||x|-|y||\geq|x|-|y|$ , in this case $x=a_n$ and $y= \frac {a_{n-1}} {z} + \ldots + \frac {a_{0}} {z^n} $

Answer 2

This is just a triangle inequality in disguise as it's equivalent to $$\left|a_n+\frac{a_{n-1}}z+\ldots+\frac{a_0}{z^n}\right|+\left|-\frac{a_{n-1}}z-\ldots-\frac{a_0}{z^n}\right|\ge|a_n|$$