Product of Complex Numbers in Polar Form/General Result

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

Let $z_j = \polar {r_j, \theta_j}$ be $z_j$ expressed in polar form for each $j \in \set {1, 2, \ldots, n}$.

Then:
 * $z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \, \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \, \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$

Let this be expressed as:
 * $\displaystyle \prod_{j \mathop = 1}^n z_j = \prod_{j \mathop = 1}^n r_j \sum_{j \mathop = 1}^n \paren {\cos \theta_j + i \sin \theta_j}$

$\map P 1$ is the case:


 * $r_1 \paren {\cos x + i \sin x} = r_1 \paren {\cos x + i \sin x }$

which is trivially true.

Basis for the Induction
$\map P 2$ is the case:


 * $z_1 z_2 = r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \, \map \sin {\theta_1 + \theta_2} }$

which is proved in Product of Complex Numbers in Polar Form.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\displaystyle \prod_{j \mathop = 1}^k z_j = \prod_{j \mathop = 1}^k r_j \sum_{j \mathop = 1}^k \paren {\cos \theta_j + i \sin \theta_j}$

Then we need to show:
 * $\displaystyle \prod_{j \mathop = 1}^{k + 1} z_j = \prod_{j \mathop = 1}^{k + 1} r_j \sum_{j \mathop = 1}^{k + 1} \paren {\cos \theta_j + i \sin \theta_j}$

Induction Step
This is our induction step:

Hence, by induction, for all $n \in \N_{>0}$:


 * $z_1 z_2 \cdots z_n = r_1 r_2 \cdots r_n \paren {\map \cos {\theta_1 + \theta_2 + \cdots + \theta_n} + i \, \map \sin {\theta_1 + \theta_2 + \cdots + \theta_n} }$