Number theory: identity of complex numbers

Question

I have seen the equation $(X^2+Y^2)(U^2+V^2) = (XU+YV)^2+(XV-YU)^2$ and I know that it is an expression of the relation $|zw| = |z| |w|$ where $z = x+yi$ and $w=u+vi$ are complex numbers.

My question is: why does the equation holds true?

Answer 1

The simplest way to show this is to use the relation $|z|^2=z\overline{z}$. Then $$|zw|^2=zw\overline{zw}=z\overline{z}w\overline{w}=|z|^2|w|^2.$$

Answer 2

We have $zw = (xu+yv) + i(xv-yu)$ and $|z|^2=x^2+y^2$, $|w|^2=u^2+y^2$.

Then $|zw|^2 =(xu+yv)^2 + (xv-yu)^2$ and $|z|^2 |w|^2 = (x^2+y^2)(u^2+v^2)$.

As you already have noted, these expressions are equal.

Answer 3

You can give it a geometric interpretation as follows assuming that $U^2+V^2 \neq 0$:

• Consider the vectors $a :=\begin{pmatrix}U \\ V \end{pmatrix}$ and $b :=\begin{pmatrix}V \\ -U \end{pmatrix}$ in the plane.
• $\Rightarrow a$ and $b$ are orthogonal, since $a\cdot b = 0$.
• Normalizing gives you two orthonormal vectors: $$o_1 =\frac{1}{|a|}a, o_2 = \frac{1}{|b|}b \mbox{ with } |a| = |b| = \sqrt{U^2+V^2}$$
• Now, you can decompose the vector $c=\begin{pmatrix}X \\ Y \end{pmatrix}$ with respect to $o_1,o_2$: $$c = (c\cdot o_1)o_1 + (c\cdot o_2)o_2$$
• Since $o_1,o_2$ are orthonormal you have $$X^2+Y^2 = |c|^2 = (c\cdot o_1)^2 + (c\cdot o_2)^2$$ $$ = \frac{(XU+YV)^2}{U^2+V^2} + \frac{(XV-YU)^2}{U^2+V^2}$$ $$\Leftrightarrow (X^2+Y^2)(U^2+V^2) = (XU+YV)^2 + (XV-YU)^2$$