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 $\theta = \map \arg z$
- $z = r e^{i \theta}$
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:
\(\ds z\) | \(=\) | \(\ds \map \Re z + i \map \Im z\) | Definition of Complex Number | |||||||||||
\(\ds \) | \(=\) | \(\ds r \cos \theta + i r \sin \theta\) | from $(1)$ and $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds r \paren {\cos \theta + i \sin \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds re^{i \theta}\) | 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:
\(\ds \cmod {r e^{i \theta} }\) | \(=\) | \(\ds \sqrt {\paren {r \cos \theta}^2 + \paren {r \sin \theta}^2}\) | Definition of Modulus of Complex Number | |||||||||||
\(\ds \) | \(=\) | \(\ds r \sqrt {\cos^2 \theta + \sin^2 \theta}\) | as $r \ge 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds r\) | 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
- 2001: Christian Berg: Kompleks funktionsteori $\S 1.5$