Product of Complex Number with Conjugate

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Theorem

Let $z = a + i b \in \C$ be a complex number.

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


Then:

$z \overline z = a^2 + b^2 = \cmod z^2$

and thus is wholly real.


Proof

By the definition of a complex number, let $z = a + i b$ where $a$ and $b$ are real numbers.

Then:

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

As $a^2 + b^2$ is wholly real, the result follows.

$\blacksquare$


Sources

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