Complex Modulus of Reciprocal of Complex Number

Theorem

Let $z \in \C$ be a complex number such that $z \ne 0$.

Let $\cmod z$ denote the complex modulus of $z$.

Then:

$\cmod {\dfrac 1 z} = \dfrac 1 {\cmod z}$

Let $z = a + i b$.

$\ds \cmod {\frac 1 z}$

$=$

$\ds \cmod {\frac {a - i b} {a^2 + b^2} }$

$\ds $

$=$

$\ds \cmod {\frac a {a^2 + b^2} + i \frac {-b} {a^2 + b^2} }$

$\ds $

$=$

$\ds \sqrt {\paren {\frac a {a^2 + b^2} }^2 + \paren {\frac {-b} {a^2 + b^2} }^2}$

$\ds $

$=$

$\ds \frac {\sqrt {a^2 + b^2} } {a^2 + b^2}$

$\ds $

$=$

$\ds \frac 1 {\sqrt {a^2 + b^2} }$

$\ds $

$=$

$\ds \frac 1 {\cmod z}$

$\blacksquare$

$\left\vert{\dfrac 1 {5 + 12 i} }\right\rvert = \dfrac 1 {13}$