Help with Proof of the Associative Property of Addition of Complex Numbers I am trying to derive a proof of the associative property of addition of complex numbers using only the properties of real numbers.
I found the following answer but was hoping someone can explain why it is correct, since I am not satisfied with it (From Using the properties of real numbers, verify that complex numbers are associative and there exists an additive inverse):

\begin{equation}
\begin{aligned}
z_1 + (z_2 + z_3) &= (a + bi) + [(c+di) + (e+fi)] \\
&= (a+bi)+[(c+e)+(di+fi)] \\
&= [a+(c+e)]+[bi+(di+fi)] \\
&= [(a+c)+e]+[(bi+di)+fi] \\
&= [(a+c)+(bi+di)]+(e+fi) \\
&= [(a+bi)+(c+di)]+(e+fi) \\
&= (z_1+z_2)+z_3
\end{aligned}
\end{equation}
I justify step 1 by the definition of complex numbers, step 2 and 3 by commutativity in R, step 4 by associativity in R, step 5 and 6 by commutativity in R.

I don't see how step 2 and 3 are commutativity in R. Wouldn't these steps require commutativity and associativity in C as well? For example, in step 2, I am presuming the author of the proof did the following rearrangement:
\begin{equation}
[(c + di) + (e + fi)] \\
[c + di + e + fi] \\
[c + e + di + fi] \\
[(c + e) + (di + fi)] \\
\end{equation}
But doesn't the following rearrangement
\begin{equation}
[(c + di) + (e + fi)] \\
[c + di + e + fi] \\
\end{equation}
require associativity in C? And doesn't the following
\begin{equation}
[c + di + e + fi] \\
[c + e + di + fi] \\
\end{equation}
require commutativity in C?
There are some other steps I am confused by as well, like step 4 only requiring associativity in R when clearly there is an associativity in C occurring on the right end of the equation.
Can anyone provide insight into why this proof is correct?
 A: Below I've annotated the proof with justifications for each equality.
$$\begin{aligned}
&\quad\ \  z_1 + (z_2 + z_3)\\[.2em] 
&= (a + bi) + [(c+di) + (e +fi)]\quad \:\!\text{by definition of a complex number} \\[.2em]
&= (a+bi)+[(c+e)+(d+f)i]\quad\,\ \text{by definition of addition in}\ \Bbb C\\[.2em]
&= [a+(c+e)]+[(b+(d+f))i]\, \quad \text{by definition of addition in}\ \Bbb C\\[.2em] 
&= [(a+c)+e]+[((b+d)+f)i] \,\quad \text{by associativity of addition in}\ \Bbb R\\[.2em]
&= [(a+c)+(b+d)i)]+(e+fi) \quad \text{by definition of addition in}\ \Bbb C\\[.2em]
&= [(a+bi)+(c+di)]+(e+fi) \quad \text{by definition of addition in}\ \Bbb C\\[.2em]
&= (z_1+z_2)+z_3
\end{aligned}$$
A: I think the key here is to differentiate between the different '$+$' we're seeing here - there are actually $3$ different kinds, $+:\Bbb R\times\Bbb R\rightarrow\Bbb R, +:\Bbb C\times\Bbb C\rightarrow\Bbb C$ and finally the '$+$' in $a+bi$.
The '$+$' in $a+bi$ is just used in representing a complex number (but there are good reasons for why '$+$' is used, as @Bill Dubuque comments and also see this question). You could perhaps consider this '$+$' as $+:\Bbb R\times\Bbb I\rightarrow\Bbb C$ to represent the complex number. For clarity, maybe it'd be more helpful if you replace it with something like $*$ just for visualisation, or even denote a complex number $a+bi$ as $(a,b)$.
The most important thing to remember here is that $+:\Bbb C\times\Bbb C\rightarrow\Bbb C$ is defined to be $(a, b)+(c, d)=(a+c, b+d)$.

The proof is rewritten as (using $(a,b)$ to represent a complex number $a+bi$ and $+_A$ for $+:A\times A\rightarrow A$):
\begin{equation}
\begin{aligned}
z_1 +_{\Bbb C} (z_2 +_{\Bbb C} z_3) &= (a, b) +_{\Bbb C} [(c, d) +_{\Bbb C} (e,f)] \quad\quad\quad\,\,\text{representation of complex numbers}\\
&= (a, c)+_{\Bbb C}(c+_{\Bbb R}e,d+_{\Bbb R}f) \,\quad\quad\quad \text{complex addition}\\
&= (a+_{\Bbb R}(c+_{\Bbb R}e),b+_{\Bbb R}(d+_{\Bbb R}f))  \quad \text{complex addition}\\
&= ((a+_{\Bbb R}c)+_{\Bbb R}e, (b+_{\Bbb R}d)+_{\Bbb R}f) \quad\text{associativity of $+$ in $\mathbb R$}\\
&= [(a+_{\Bbb R}c,b+_{\Bbb R}d)]+_{\Bbb C}(e,f) \,\,\,\quad\quad\text{complex addition}\\
&= [(a,b)+_{\Bbb C}(c,d)]+_{\Bbb C}(e,f) \quad\quad\quad\,\text{complex addition}\\
&= (z_1+_{\Bbb C}z_2)+_{\Bbb C}z_3 \,\,\quad\quad\quad\quad\quad\quad\,\,\,\text{representation of complex numbers}
\end{aligned}
\end{equation}
A: The person who did that solution just made a mess of it.  S/he seems to be believing we can treat the imaginary $bi$ and real components $a$ and the imaginary unit $i$ as real summands and rearrange them and distribute them.  We can.  But that has to be proven.
A better proof:
$z_1 + (z_2 + z_3) = \{a + bi)\} + (\{c+di\} +\{e+fi\})$ Notation of complex numbers.  (Am adding unconventional $\{\}$ brackets to clarify what exactly are the inseparable "units" of complex numbers.  At this time we cannot break the two compenents of $\{a + bi\}$ into $a$ and $bi$ and shuffle them about as though they were real summands.)
$= \{a + bi\} + \{(c+e) + (d+f)i\}$ Definition of complex addition.  (Note we can't, at this time treat $di$ and $fi$ as components and so $(c + di) + (e+fi) = (c+e) + (di+fi)$ via commutivity of addition on reals.  Yet.  We don't have that the "+" in $c + di$ actually means $di$ added to $c$ in the same sense that $c + e$ has $e$ added to $c$ would mean.
$= \{(a + (c+e)) + (b + (d+f))i\}$ Definition of complex addition.
$=\{((a+c) + e) + ((b + d) +f\}i\}$ Associativity of addition in the reals.
$=\{(a+c) + (b+d)i\} + \{e + fi\}$ Definition of addition of complex numbers. (but going in the opposite direction.  $\{a + bi\} + \{c+di\} = \{(a+b) + (b+d)i\}$ is summands to sum.  This is the other direction: $\{(a+b) + (b+d)i\}= \{a + bi\} + \{c+di\}  $... sum to summands.)
$= (\{a + bi\} + \{c + di\}) + \{e + fi\}$ Definition of complex addition.
$= (z_1 + z_2) + z_3$.  Notation.
A better way, would be to simply not use $a + bi$ notation at all but define a complex number as an ordered pair $a + bi = (a,b)$ and $\{a + bi\} + \{c + di\} = (a,b) + (c,d) := (a+c, b+d)$.
Then the argument is:
$z_1 + [z_2 + z_3]= (a,b) + [(c,d) + (e,f)]$ Notation
$(a,b) + (c+e,d+f)$ Def.  Addition.
$(a + [c+e], b + [d+f])$ Def. Addition.
$([a+c] + e, [b+d]+f)$ Assoc. of Addition on reals.
$(a+c, b+d) + (e,f)$ Def. Addition.
$[(a,b) + (c,d)] + (e,f)$ Def. Addition.
$[z_1+ z_2] + z_3$ Notation.
Associativity of multiplication is similar but harder:
$(a,b)[(c,d)*(e,f)] = $
$(a,b)(ce - df, de + cf) =$
$(a(ce - df) - b(de + cf), b(ce - df) + a(de+cf))=$
$(ace - adf -bde - bcf, ade + acf + bce - bdf)$.
And $[(a,b)*(c,d)]*(e,f)=$
$(ac - bd, bc + ad)(e,f) = $
$((ac - bd)e - (bc + ad)f, (bc+ad)e + (ac-bd)f) = $
$(ace - bde - bcf + adf, ade + afc +bce = bdf)=(a,b)[(c,d)*(e,f)]$
