On page #191 of the text: A Decade of the Berkeley Math Circle, by Zvezdelina Stankova, Tom Rike; as shown in Exercise 12, it is given the complex multiplication based interpretation of the above, i.e.:
$(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2$ for real $a,b,c,d$.
It talks about two facts:
(i) conjugation preserves modulus,
(ii) modulus respects complex multiplication.
It is easy to see that given two complex numbers $c_1 = a+ib, c_2 = c+id$, on doing multiplication leads to $(ac-bd)+i(ad+bc)$.
The modulus of the two complex numbers $c_1, c_2$ is given by $\sqrt{a^2 + b^2}, \sqrt{c^2+d^2}$, while of their product is given by : $\sqrt{(ac-bd)^2 + (ad+bc)^2}$.
This is the use of second fact, as $|c_1\cdot c_2|^2 = |c_1|^2|c_2|^2$, as shown below:
$(\sqrt{a^2 + b^2} \sqrt{c^2+d^2})^2 = (a^2 + b^2)(c^2+d^2)= a^2c^2+ a^2d^2+b^2c^2+b^2d^2$,
while: $(ac-bd)^2+(ad+bc)^2 = a^2c^2+b^2d^2-2abcd + a^2d^2+b^2c^2+2abcd= a^2c^2+b^2d^2+a^2d^2+b^2c^2$
But, can not see how the first fact is being used, i.e. conjugation preserves modulus.
Edit-- As per the comment of @dxiv can interpret that taking the conjugates of the complex numbers $c_1$ will still result in the same equality being observed. $\overline c_1=(a-ib)$, with modulus being still : $\sqrt{a^2+b^2}$, also the product of $\overline c_1c_2 = (a-ib)(c+id) = (ac +bd)+i(ad-bc)$.
If however conjugates of both $c_1, c_2$ are taken, then: $\overline c_1=(a-ib), \overline c_2=(c-id)$. Now, $\overline c_1\cdot \overline c_2= (a-ib)(c-id)= (ac-bd)-i(ad+bc)$. The modulus of $\overline c_1= \sqrt{a^2+b^2}$, and also the modulus of $\overline c_2$ is preserved. Also, the modulus of the product of conjugates is : $\sqrt{(ac-bd)^2+(ad+bc)^2}$.
On squaring the above two modulus, get :
l.h.s. :$|\overline c_1|^2|\overline c_2|^2 = (a^2+b^2)(c^2+d^2)= a^2c^2+a^2d^2+b^2c^2+b^2d^2$
r.h.s.: $(\overline c_1\cdot \overline c_2)^2=(ac-bd)^2+(ad+bc)^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$
