Modulus and Argument of Complex Exponential

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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 $\theta = \map \arg z$

if and only if:

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

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 $\hointr a {a + 2 \pi}$.


Proof

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:

\(\displaystyle z\) \(=\) \(\displaystyle \map \Re z + i \map \Im z\) Definition of Complex Number
\(\displaystyle \) \(=\) \(\displaystyle r \cos \theta + i r \sin \theta\) from $(1)$ and $(2)$
\(\displaystyle \) \(=\) \(\displaystyle r \paren {\cos \theta + i \sin \theta}\)
\(\displaystyle \) \(=\) \(\displaystyle re^{i \theta}\) Euler's Formula

$\Box$


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:

\(\displaystyle \cmod {r e^{i \theta} }\) \(=\) \(\displaystyle \sqrt {\paren {r \cos \theta}^2 + \paren {r \sin \theta}^2}\) Definition of Modulus of Complex Number
\(\displaystyle \) \(=\) \(\displaystyle r \sqrt {\cos^2 \theta + \sin^2 \theta}\) as $r \ge 0$
\(\displaystyle \) \(=\) \(\displaystyle r\) Sum of Squares of Sine and Cosine

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:

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

$\blacksquare$


Sources