Exponential of Sum/Complex Numbers/General Result

Theorem
Let $m \in \N_{>0}$ be a natural number.

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

Let $\exp z$ be the exponential of $z$.

Then:


 * $\displaystyle \map \exp {\sum_{j \mathop = 1}^m z_j} = \prod_{j \mathop = 1}^m \paren {\exp z_j}$

Proof
The proof proceeds by induction.

For all $m \in \N_{>0}$, let $\map P m$ be the proposition:
 * $\displaystyle \map \exp {\sum_{j \mathop = 1}^m z_j} = \prod_{j \mathop = 1}^m \paren {\exp z_j}$

$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

Basis for the Induction
$\map P 2$ is the case:
 * $\map \exp {z_1 + z_2} = \paren {\exp z_1} \paren {\exp z_2}$

This is proved in Exponential of Sum: Complex Numbers.

Thus $\map P 2$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $\displaystyle \map \exp {\sum_{j \mathop = 1}^k z_j} = \prod_{j \mathop = 1}^k \paren {\exp z_j}$

from which it is to be shown that:
 * $\displaystyle \map \exp {\sum_{j \mathop = 1}^{k + 1} z_j} = \prod_{j \mathop = 1}^{k + 1} \paren {\exp z_j}$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall m \in \N_{>0}: \displaystyle \map \exp {\sum_{j \mathop = 1}^m z_j} = \prod_{j \mathop = 1}^m \paren {\exp z_j}$