Modulus and Argument of Complex Exponential

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

Let $\left[{a \,.\,.\, a + 2 \pi}\right)$ be a half open interval of length $2 \pi$.

Let $r \in \left[{0 \,.\,.\, \infty}\right)$ and $\theta \in \left[{a \,.\,.\, a + 2 \pi}\right)$.

Then $r = \left\vert{z}\right\vert$ and $\theta = \arg \left({z}\right)$ iff $z = re^{i \theta}$.

Here, $\left\vert{z}\right\vert$ denotes the modulus of $z$, $\arg \left({z}\right)$ denotes the argument of $z$, and $e$ is the complex exponential function.

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

Necessary condition
Suppose that $r = \left\vert{z}\right\vert$.

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

Suppose $z \ne 0$ and $\theta = \arg \left({z}\right)$.

By definition of argument, the following two equations hold:


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

where $\operatorname{Re} \left({z}\right)$ denotes the real part of $z$, and $\operatorname{Im} \left({z}\right)$ denotes the imaginary part of $z$.

Then:

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

From the equations above, we find:


 * $\operatorname{Re} \left({re^{i \theta} }\right) = r \cos \theta \ ,$
 * $\operatorname{Im} \left({re^{i \theta} }\right) = r \sin \theta$

Then:

If $r \ne 0$, we find $\arg \left({re^{i \theta} }\right)$ by solving the two equations from definition of argument:


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

We find:


 * $\arg \left({re^{i \theta} }\right) = \theta$