I am trying to prove complex distributivity.
The proofs that I have seen proceed as follows:
$z_1(z_2 + z_3) = (a + ib)[(c + id) + (e + if)]$
$= a+ib.(c+e)+i(d+f)]$
$= a.(c+e)-b.(d+f)+i[a.(d+f)+b(c+e)]$
$= ac+ae-bd-bf+i[ad+af+bc+be]$ (1)
$= [(ac-bd)+i(ad+bc)]+[(ae-bf)+i(af+be)]$ (2)
$= z_1 z_2 + z_2 z_3$
My proofs proceed in the same way, but I get stuck at (1), because it doesn't seem to me that it is valid to use the operation necessary to get to (2). In going from (1) to (2), $i$ must first be distributed through $[ad+af+bc+be]$, which, as I understand it, is using the complex distributive property (since $i = 0 + 1i$ is a complex number), which is precisely what we're trying to prove? You then have $= ac + ae - bd - bf + i ad + i af + i bc + i be$, which then requires the using commutativity of complex addition to get into the form $[(ac-bd)+i(ad+bc)]+[(ae-bf)+i(af+be)]$
Am I right here? And if so, then how should the proof actually proceed, because I'm stuck at this point. If not, then why am I wrong (why is this valid)?
I would greatly appreciate it if people could please take the time to clarify this.