2016/121) Express $(1+a^2)(1+b^2)$ as sum of 2 squares

we have $(1+a^2)(1+b^2) = (1+ ia)(1-ia)(1+ib)(1-ib)$ factoring over complex
$=((1+ia)(1+ib))(1-ia)(1-ib))$
$=(( 1 - ab) + i(a+b))((1 - ab) - i(a+b)) = (1- ab)^2 + (a+b)^2$ multiplication over complex