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?