Modulus and Argument of Complex Exponential

Theorem
Let $z \in \C$ be a complex number.

Let $\hointr a {a + 2 \pi}$ be a half open interval of length $2 \pi$.

Let $r \in \hointr 0 {+\infty}$ and $\theta \in \hointr a {a + 2 \pi}$.

Then:
 * $r = \cmod z$ and $\theta = \map \arg z$


 * $z = r e^{i \theta}$
 * $z = r e^{i \theta}$

where:
 * $\cmod z$ denotes the modulus of $z$
 * $\map \arg z$ denotes the argument of $z$
 * $x \mapsto e^x$ is the complex exponential function.

If $z = 0$ or $r = 0$, then $\theta$ may be any number in $\hointr a {a + 2 \pi}$.

Necessary condition
Let $r = \cmod z$.

If $z = 0$, we have:
 * $z = 0e^{i \theta} = re^{i \theta}$

Suppose $z \ne 0$ and $\theta = \map \arg z$.

By definition of argument, the following two equations hold:


 * $(1): \quad \dfrac {\map \Re z} r = \cos \theta$
 * $(2): \quad \dfrac {\map \Im z} r = \sin \theta$

where:
 * $\map \Re z$ denotes the real part of $z$
 * $\map \Im z$ denotes the imaginary part of $z$.

Then:

Sufficient condition
Let $z = re^{i \theta}$.

From the equations above, we find:


 * $\map \Re {re^{i \theta} } = r \cos \theta$
 * $\map \Im {re^{i \theta} } = r \sin \theta$

Then:

If $r \ne 0$, we find $\map \arg {r e^{i \theta} }$ by solving the two equations by definition of argument:


 * $(1): \quad \dfrac {r \cos \theta} r = \map \cos {\map \arg {r e^{i \theta} } }$
 * $(2): \quad \dfrac {r \sin \theta} r = \map \sin {\map \arg {r e^{i \theta} } }$

We find:


 * $\map \arg {r e^{i \theta} } = \theta$