Complex Modulus equals Complex Modulus of Conjugate

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Theorem

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

Let $\overline z$ denote the complex conjugate of $z$.

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


Then:

$\cmod z = \cmod {\overline z}$


Proof

Let $z = a + b i$.

Then:

\(\ds \cmod z\) \(=\) \(\ds \cmod {a + b i}\) Definition of $z$
\(\ds \) \(=\) \(\ds \sqrt {a^2 + b^2}\) Definition of Complex Modulus

and:

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

$\blacksquare$


Sources

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