Uniquely determining complex multiplication I encountered the following problem and it's not entirely clear to me exactly what I am supposed to do:

Show that the following rules uniquely determine complex multiplication on $\mathbb{C}=\mathbb{R^2}$:
  
  
*
  
*(a) $(z_1+z_2)w=z_1w+z_2w$
  
*(b) $z_1z_2=z_2z_1$
  
*(c) $i\cdot i=-1$
  
*(d) $z_1(z_2z_3)=(z_1z_2)z_3$
  
*(e) If $z_1$ and $z_2$ are real, $z_1\cdot z_2$ is the usual product of real numbers.
  

I get that (a) refers to distributivity, (b) refers to commutativity, (c) I'm not sure what the goal is there (is there a special name for this property?), (d) refers to associativity, and (e) refers to something else. 
Looking at (b), for example (with $z_1=a_1+b_1i$ and $z_2=a_2+b_2i$), I get that
\begin{align}
z_1z_2 &= (a_1+b_1i)(a_2+b_2i)\\[0.5em]
&= (a_1a_2+a_1b_2i+b_1ia_2+b_1ib_2i)\\[0.5em]
&= (a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i
\end{align}
and
\begin{align}
z_2z_1 &= (a_2+b_2i)(a_1+b_1i)\\[0.5em]
&= (a_2a_1+a_2b_1i+b_2ia_1+b_2ib_1i)\\[0.5em]
&= (a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i
\end{align}
Since $\operatorname{Re}(z_1z_2)=\operatorname{Re}(z_2z_1)$ and $\operatorname{Im}(z_1z_2)=\operatorname{Im}(z_2z_1)$, it is clear that $z_1z_2=z_2z_1$. But what does this really show? It seems like the question is asking for more (something about uniqueness, etc.). 
Any ideas?
 A: $$z_1 z_2 = (a_1 + b_1 i)(a_2 + b_2 i) = \text{(rule a)}$$
$$ a_1(a_2 + b_2 i) + (b_1 i)(a_2 + b_2 i) = \text{(rule b)}$$
$$ (a_2 + b_2 i) a_1 + (a_2 + b_2 i)(b_1 i) = \text{(rule a)}$$
$$ a_2 a_1 + (b_2 i) a_1 + a_2 (b_1 i) + (b_2 i)(b_1 i) = \text{(rule d)}$$
$$ a_2 a_1 + (b_2 a_1) i + (a_2 b_1) i + ((b_2 i)b_1) i = \text{(rule b)}$$
$$ a_2 a_1 + i(b_2 a_1) + i(a_2 b_1) + ((i b_2)b_1) i = \text{(rule a and d)}$$
$$ a_2 a_1 + i(b_2 a_1 + a_2 b_1) + (i (b_2 b_1)) i = \text{(rule b)}$$
$$ a_2 a_1 + (b_2 a_1 + a_2 b_1)i + ((b_2 b_1) i) i = \text{(rule d)}$$
$$ a_2 a_1 + (b_2 a_1 + a_2 b_1)i + (b_2 b_1) (i i) = \text{(rule c)}$$
$$ a_2 a_1 + (b_2 a_1 + a_2 b_1)i + (b_2 b_1) (-1) = \text{(rule e)}$$
$$ a_2 a_1 - b_1 b_2 + (b_2 a_1 + a_2 b_1)i $$
A: Well, using only the rules given, denoting the complex product with $\cdot$, we can say that for any complex numbers $z = a + i \cdot b$ and $w = c + i\cdot d$, we must have (using $(d)$ whenever there are terms involving at least two multiplications and writing $a \cdot b \cdot c$ for $a \cdot (b \cdot c) = (a \cdot b) \cdot c)$):
$$\begin{align*} z\cdot w &= (a+i\cdot b)\cdot (c+i\cdot d) \\ &= a \cdot (c+i\cdot d) + i\cdot b \cdot (c+i\cdot d) \quad \text{using (a)} \\ &= (c+i\cdot d) \cdot a + (c+i\cdot d)\cdot i\cdot b \quad \text{using (b)} \\ &= c\cdot a + i\cdot d \cdot a + c \cdot i\cdot b + i\cdot d \cdot i\cdot b \quad \text{using (a)} \\ &= c \cdot a + i \cdot d \cdot a + i \cdot c \cdot b + i \cdot i \cdot d \cdot b \quad \text{using (b)} \\ &= c \cdot a + i \cdot d \cdot a + i \cdot c \cdot b  - d \cdot b \quad \text{using (c)} \\ &= c \cdot a - d \cdot b + i \cdot (d \cdot a + c \cdot b)\quad \text{using (a)} \\
&= c a - d b + i \cdot (d a + c b)\quad \text{using (e), denoting real multiplication by juxtaposition} \\
\end{align*}$$
So complex multiplication is uniquely determined by these rules.
