Triangle Inequality/Complex Numbers/General Result

Theorem
Let $z_1, z_2, \ldots, z_n \in \C$ be complex numbers.

Let $\left\vert{z}\right\vert$ be the modulus of $z$.

Then:
 * $\left\vert{z_1 + z_2 + \cdots + z_n}\right\vert \le \left\vert{z_1}\right\vert + \left\vert{z_2}\right\vert + \cdots + \left\vert{z_n}\right\vert$

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\left\vert{z_1 + z_2 + \cdots + z_n}\right\vert \le \left\vert{z_1}\right\vert + \left\vert{z_2}\right\vert + \cdots + \left\vert{z_n}\right\vert$

$P \left({1}\right)$ is trivially true:
 * $\left\vert{z_1}\right\vert \le \left\vert{z_1}\right\vert$

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $\left\vert{z_1 + z_2}\right\vert \le \left\vert{z_1}\right\vert + \left\vert{z_2}\right\vert$

which has been proved in Triangle Inequality: Complex Numbers.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\left\vert{z_1 + z_2 + \cdots + z_k}\right\vert \le \left\vert{z_1}\right\vert + \left\vert{z_2}\right\vert + \cdots + \left\vert{z_k}\right\vert$

Then we need to show:
 * $\left\vert{z_1 + z_2 + \cdots + z_{k+1} }\right\vert \le \left\vert{z_1}\right\vert + \left\vert{z_2}\right\vert + \cdots + \left\vert{z_{k+1} }\right\vert$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N_{>0}: \left\vert{z_1 z_2 \cdots z_n}\right\vert = \left\vert{z_1}\right\vert \cdot \left\vert{z_2}\right\vert \cdots \left\vert{z_n}\right\vert$