Complex Modulus of Reciprocal of Complex Number
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Theorem
Let $z \in \C$ be a complex number such that $z \ne 0$.
Let $\cmod z$ denote the complex modulus of $z$.
Then:
- $\cmod {\dfrac 1 z} = \dfrac 1 {\cmod z}$
Proof
Let $z = a + i b$.
\(\ds \cmod {\frac 1 z}\) | \(=\) | \(\ds \cmod {\frac {a - i b} {a^2 + b^2} }\) | Reciprocal of Complex Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {\frac a {a^2 + b^2} + i \frac {-b} {a^2 + b^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\paren {\frac a {a^2 + b^2} }^2 + \paren {\frac {-b} {a^2 + b^2} }^2}\) | Definition of Complex Modulus | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt {a^2 + b^2} } {a^2 + b^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt {a^2 + b^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\cmod z}\) | Definition of Complex Modulus |
$\blacksquare$
Examples
Complex Modulus of $\dfrac 1 {5 + 12 i}$
- $\left\vert{\dfrac 1 {5 + 12 i} }\right\rvert = \dfrac 1 {13}$