# Exponential of Sum/Complex Numbers

## Theorem

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

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

Then:

$\map \exp {z_1 + z_2} = \paren {\exp z_1} \paren {\exp z_2}$

### Corollary

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

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

Then:

$\map \exp {z_1 - z_2} = \dfrac {\exp z_1} {\exp z_2}$

### General Result

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:

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

## Proof

This proof is based on the definition of the complex exponential as the unique solution of the differential equation:

$\dfrac \d {\d z} \exp = \exp$

which satisfies the initial condition $\map \exp 0 = 1$.

Define the complex function $f: \C \to \C$ by:

$\map f z = \map \exp z \, \map \exp {z_1 + z_2 - z}$

Then find its derivative:

 $\ds D_z \, \map f z$ $=$ $\ds \paren {D_z \, \map \exp z} \map \exp {z_1 + z_2 - z} + \map \exp z \paren {D_z \map \exp {z_1 + z_2 - z} }$ Derivative of Complex Composite Function $\ds$ $=$ $\ds \map \exp z \, \map \exp {z_1 + z_2 - z} + \map \exp z \, \map \exp {z_1 + z_2 - z} \map {D_z} {z_1 + z_2 - z}$ as $\exp$ is its own derivative $\ds$ $=$ $\ds \map \exp z \, \map \exp {z_1 + z_2 - z} - \map \exp z \, \map \exp {z_1 + z_2 - z}$ Derivative of Complex Power Series $\ds$ $=$ $\ds 0$

From Zero Derivative implies Constant Complex Function, it follows that $f$ is constant.

Then:

 $\ds \map \exp {z_1} \, \map \exp {z_2}$ $=$ $\ds \map f {z_1}$ $\ds$ $=$ $\ds \map f 0$ as $f$ is constant $\ds$ $=$ $\ds \map \exp 0 \, \map \exp {z_1 + z_2}$ $\ds$ $=$ $\ds \map \exp {z_1 + z_2}$ as $\map \exp 0 = 1$

$\blacksquare$