Exponential of Sum/Complex Numbers

From ProofWiki
Jump to navigation Jump to search

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 $\map \exp 0 = 1$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Exponential_of_Sum/Complex_Numbers&oldid=396485"