Absolute Value of Complex Dot Product is Commutative

Theorem

Let $z_1$ and $z_2$ be complex numbers.

Let $z_1 \circ z_2$ denote the (complex) dot product of $z_1$ and $z_2$.

Then:

$\size {z_1 \circ z_2} = \size {z_2 \circ z_1}$

where $\size {\, \cdot \,}$ denotes the absolute value function.

Proof

From Dot Product Operator is Commutative‎:

$z_1 \circ z_2 = z_2 \circ z_1$

The result follows trivially.

$\blacksquare$

Examples

Example: $\size {\paren {2 + 5 i} \circ \paren {3 - i} } = \size {\paren {3 - i} \circ \paren {2 + 5 i} }$

Example: $\size {\paren {2 + 5 i} \circ \paren {3 - i} }$

Let:

$z_1 = 2 + 5 i$

$z_2 = 3 - i$

Then:

$\size {z_1 \circ z_2} = 1$

Example: $\size {\paren {3 - i} \circ \paren {2 + 5 i} }$

Let:

$z_1 = 3 - i$

$z_2 = 2 + 5 i$

Then:

$\size {z_1 \circ z_2} = 1$

As can be seen:

$\size {\paren {2 + 5 i} \circ \paren {3 - i} } = \size {\paren {3 - i} \circ \paren {2 + 5 i} }$

$\blacksquare$