Proving a surjective homomorphism from $a+bi \to a^2+b^2$

Question

I am trying to prove the surjective homomorphism between $f: \mathbb{C}^\times \to \mathbb{R}^\times_{>0}$ of $a+bi \to a^2 + b^2$

Currently this is my working out to show that it is a homomorphism, approached from both the left hand and the right hand side to show that $f(z_1z_2) = f(z_1)f(z_2)$

But it appears that I have made a mistake somewhere, perhaps fundamentally as I cannot find a calculative error.

Note for $z_1 = a_1 + b_1 i, z_2 = a_2 + b_2i \in \mathbb{C}^{\times}$

$f(z_1 z_2) = f(a_1a_2 + a_1b_2i+b_1a_2i -b_1b_2) = (a_1a_2 -b_1b_2)^2 +(a_1b_2i + b_1a_2i)^2$

$=a_1^2a_2^2 - 2a_1a_2b_1b_2 + b_1^2b_2^2 - a_1^2b_2^2 - 2a_1b_2b_1a_2i - b_1^2a_2^2$

$=????$

$= a_1^2a_2^2+a_1^2b_2^2+b_1^2a_2^2+b_1^2b_2^2$

$= (a_1^2 + b_1^2)(a_2^2 + b_2^2)$

$=f(z_1)f(z_2)$

Thanks

Answer 1

Correction: $f(z_1z_2) = (a_1a_2-b_1b_2)^2 + (a_1b_2+b_2a_1)^2$. Expand from here.