Modulus of Product

Theorem

Let $z_1, z_2 \in \C$ be complex numbers.

Let $\left\vert{z}\right\vert$ be the modulus of $z$.

Then $\left\vert{z_1 z_2}\right\vert = \left\vert{z_1}\right\vert \cdot \left\vert{z_2}\right\vert$.

Corollary

Let $x, y \in \R$ be real numbers.

Let $\left\vert{x}\right\vert$ be the absolute value of $x$.

Then $\left\vert{xy}\right\vert = \left\vert{x}\right\vert \cdot \left\vert{y}\right\vert$.

Proof

Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.

	\(\displaystyle \)	\(\displaystyle \left\vert{z_1 z_2}\right\vert\)	\(=\)	\(\displaystyle \sqrt {\left({x_1 x_2 - y_1 y_2}\right)^2 + \left({x_1 y_2 + x_2 y_1}\right)^2}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt {\left({x_1^2 x_2^2 + y_1^2 y_2^2 - 2 x_1 x_2 y_1 y_2}\right) + \left({x_1^2 y_2^2 + x_2^2 y_1^2 + 2 x_1 x_2 y_1 y_2}\right)}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt {x_1^2 x_2^2 + y_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2}\)	\(\displaystyle \)

	\(\displaystyle \)	\(\displaystyle \left\vert{z_1}\right\vert \cdot \left\vert{z_2}\right\vert\)	\(=\)	\(\displaystyle \sqrt {x_1^2 + y_1^2} \sqrt {x_2^2 + y_2^2}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt {\left({x_1^2 + y_1^2}\right) \left({x_2^2 + y_2^2}\right)}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt {x_1^2 x_2^2 + y_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2}\)	\(\displaystyle \)

$\blacksquare$

Proof of Corollary

Follows directly from the fact that the Multiplicative Group of Reals Subgroup of Complex.

Therefore, any result applying to all complex numbers will also hold for all reals.

Alternatively, a direct approach can be taken:

	\(\displaystyle \)	\(\displaystyle \left\vert{xy}\right\vert\)	\(=\)	\(\displaystyle \sqrt {\left({xy}\right)^2}\)	\(\displaystyle \)	Definition of absolute value
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt {x^2y^2}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sqrt {x^2} \sqrt{y^2}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert{x}\right\vert \cdot \left\vert{y}\right\vert\)	\(\displaystyle \)

$\blacksquare$

Alternatively, as Real Numbers form Ordered Integral Domain, we can go straight to Product of Absolute Values and use that directly.

$\blacksquare$‎

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.16$