Reciprocal of Complex Number in terms of Conjugate and Modulus

From ProofWiki
Jump to navigation Jump to search

Theorem

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

The reciprocal of $z$ can be expressed as:

$\dfrac 1 z = \dfrac {\overline z} {\cmod z^2}$

where:

$\overline z$ denotes the complex conjugate of $z$
$\cmod z^2$ denotes the modulus of $z$.


Proof

Let $z$ be defined as:

$z = a = i b$


Then:

\(\ds \dfrac 1 z\) \(=\) \(\ds \dfrac {a - i b} {a^2 + b^2}\) Reciprocal of Complex Number
\(\ds \) \(=\) \(\ds \dfrac {a - i b} {\paren {\sqrt {a^2 + b^2} }^2}\) rearranging
\(\ds \) \(=\) \(\ds \dfrac {\overline z} {\paren {\sqrt {a^2 + b^2} }^2}\) Definition of Complex Conjugate
\(\ds \) \(=\) \(\ds \dfrac {\overline z} {\cmod z^2}\) Definition of Complex Modulus

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Reciprocal_of_Complex_Number_in_terms_of_Conjugate_and_Modulus&oldid=633278"