What is a point of using complex numbers here?

[Hill]

Homework Statement

"Here is a basic fact about integers that has many uses in number theory: If two integers can be expressed as the sum of two squares, then so can their product. With the understanding that each symbol denotes an integer, this says that if $M = a^2 + b^2$ and $N = c^2 + d^2$, then $MN = p^2 + q^2$. Prove this result by considering $|(a + ib)(c + id)|^2$."

Relevant Equations

$|x+iy|^2 = x^2 + y^2$

Firstly, the exercise itself is not difficult:
On one hand, $$|(a + ib)(c + id)|^2 = |a + ib|^2|c + id|^2 = (a^2 + b^2) (c^2 + d^2) = MN.$$
On the other hand, $(a + ib)(c + id) = p+ iq$ for some integers p and q, and so $$|(a + ib)(c + id)|^2 = |p + iq|^2 = p^2 + q^2.$$
Thus, $MN = p^2 + q^2.$

However, it can be done quite straightforwardly, without considering the complex numbers, e.g.,
$$MN = (a^2 + b^2) (c^2 + d^2) = a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 =$$$$a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 + 2abcd - 2abcd = $$$$(a^2 c^2 + 2abcd + b^2 d^2) + (a^2 d^2 - 2abcd + b^2 c^2) =$$$$(ac + bd)^2 + (ad - bc)^2 = p^2 + q^2.$$

I don't see an advantage of considering the complex numbers in this case. What am I missing?

 

[PeroK]

You could prove this slightly more easily by using the properties of the complex modulus. The required algebra is already encapsulated in the equation $|zw|^2 = |z|^2|w|^2$.

 

[Hill]

You could prove this slightly more easily by using the properties of the complex modulus. The required algebra is already encapsulated in the equation $|zw|^2 = |z|^2|w|^2$.

Thank you. I thought I've used it in this line:
$$|(a + ib)(c + id)|^2 = |a + ib|^2|c + id|^2 = (a^2 + b^2) (c^2 + d^2) = MN.$$
Is there something else there that I didn't use and that could simplify it further?

 

[PeroK]

Thank you. I thought I've used it in this line:
$$|(a + ib)(c + id)|^2 = |a + ib|^2|c + id|^2 = (a^2 + b^2) (c^2 + d^2) = MN.$$
Is there something else there that I didn't use and that could simplify it further?

That looks simpler than the alternative to me.