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:

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

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:

\(\displaystyle z \overline z\) \(=\) \(\displaystyle \left({a + i b}\right) \left({a - i b}\right)\) Definition of Complex Conjugate
\(\displaystyle \) \(=\) \(\displaystyle a^2 + a \cdot i b + a \cdot \left({-i b}\right) + i \cdot \left({-i}\right) \cdot b^2\) Definition of Complex Multiplication
\(\displaystyle \) \(=\) \(\displaystyle a^2 + i a b - i a b + b^2\)
\(\displaystyle \) \(=\) \(\displaystyle a^2 + b^2\)

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

$\blacksquare$


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