Exponential on Complex Plane is Group Homomorphism

Theorem
Let $\left({\C, +}\right)$ be the additive group of complex numbers.

Let $\left({\C_{\ne 0}, \times}\right)$ be the multiplicative group of complex numbers.

Let $\exp: \left({\C, +}\right) \to \left({\C_{\ne 0}, \times}\right)$ be the mapping:
 * $x \mapsto \exp \left({x}\right)$

where $\exp$ is the complex exponential function.

Then $\exp$ is a group homomorphism.

Proof
If $z \in \C$, then by the definition of the complex exponential function, $\exp$ is a mapping $\C \to \C_{\ne 0}$.

By the Exponent of Sum property, if $z_1, z_2 \in \C$, then:
 * $\exp \left({z_1 + z_2}\right) = \exp \left({z_1}\right) \exp \left({z_2}\right)$

Therefore, $\exp: \left({\C, +}\right) \to \left({\C_{\ne 0}, \times}\right)$ is a group homomorphism.