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?

  • $\begingroup$ (c) tells us that i exists in this expanded number system, and is the squreroot of -1. This is used to show that $b_1 i b_2 i = - b_1 b_2$ $\endgroup$ – mlu Dec 22 '15 at 23:03
  • $\begingroup$ Right, but I'm not sure how that gets me closer to the goal of proving that those five rules together uniquely determine multiplication in $\mathbb{C}$. I guess the real question is: What does the author mean by "uniquely determine"? And that's why I have come here...for suggestions. I really have no clue. $\endgroup$ – interrogative Dec 22 '15 at 23:05
  • $\begingroup$ Using only these rules, and nothing else you know about complex numbers, you can calculate the product of two complex numbers. That is how they uniquely determines the complex multiplication. $\endgroup$ – mlu Dec 22 '15 at 23:08
  • $\begingroup$ @mlu That seems plausible but I am slightly confused about one thing--why would I need (c) to show (b) when it seems like I am supposed to proceed in a linear fashion? Also, why would (d) need to be included if I am only worried about computing the product of two complex numbers? And the last problem doesn't make a great deal of sense to me. And for (a), I'm guessing $w\in\mathbb{C}$, but it seems like it's implicit that $w\in\mathbb{R}$. $\endgroup$ – interrogative Dec 22 '15 at 23:11
  • $\begingroup$ If one gives you $\star:\mathbb{C}\times\mathbb{C}\rightarrow\mathbb{C}$ a law that satisfies $(a),(b),(c),(d)$ and $(e)$, then $\star$ is the complex multiplication. This is what you have to prove by taking $z_1:=a_1+ib_1,z_2:=a_2+ib_2\in\mathbb{C}$ and computing $z_1\star z_2$. Using only $(a),(b),(c),(d)$ and $(e)$, you will find: $$z_1\star z_2=(a_1a_2-b_1b_2)+i(a_1b_2+b_1a_2).$$ Notice that you will use ALL the assumptions $(a),(b),(c),(d)$ and $(e)$ on $\star$. $\endgroup$ – C. Falcon Dec 22 '15 at 23:18

$$z_1 z_2 = (a_1 + b_1 i)(a_2 + b_2 i) = \text{(rule a)}$$ $$ a_1(a_2 + b_2 i) + (b_1 i)(a_2 + b_2 i) = \text{(rule b)}$$ $$ (a_2 + b_2 i) a_1 + (a_2 + b_2 i)(b_1 i) = \text{(rule a)}$$ $$ a_2 a_1 + (b_2 i) a_1 + a_2 (b_1 i) + (b_2 i)(b_1 i) = \text{(rule d)}$$ $$ a_2 a_1 + (b_2 a_1) i + (a_2 b_1) i + ((b_2 i)b_1) i = \text{(rule b)}$$ $$ a_2 a_1 + i(b_2 a_1) + i(a_2 b_1) + ((i b_2)b_1) i = \text{(rule a and d)}$$ $$ a_2 a_1 + i(b_2 a_1 + a_2 b_1) + (i (b_2 b_1)) i = \text{(rule b)}$$ $$ a_2 a_1 + (b_2 a_1 + a_2 b_1)i + ((b_2 b_1) i) i = \text{(rule d)}$$ $$ a_2 a_1 + (b_2 a_1 + a_2 b_1)i + (b_2 b_1) (i i) = \text{(rule c)}$$ $$ a_2 a_1 + (b_2 a_1 + a_2 b_1)i + (b_2 b_1) (-1) = \text{(rule e)}$$ $$ a_2 a_1 - b_1 b_2 + (b_2 a_1 + a_2 b_1)i $$

  • $\begingroup$ I like the flow of this but the side comments are a little confusing to me...it appears you are using them to say, "This rule applies to get what follows." For example, in the first line you reference (a) even though the product is simply by the definition of complex numbers but you do use (a) to get from the first to the second line. And so on and so forth. Does that make sense? Other than that, makes good sense to me! $\endgroup$ – interrogative Dec 22 '15 at 23:40
  • $\begingroup$ the rules given are the one used to get to the next line (thats why they are placed between the = and the next line, perhaps they should be above the = sign) . $\endgroup$ – mlu Dec 22 '15 at 23:43
  • $\begingroup$ As an aside, have you ever considered using the align environment or the \tag feature? Pretty handy for those kinds of things. Anyway, thanks for the help! $\endgroup$ – interrogative Dec 22 '15 at 23:46

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.

  • $\begingroup$ Interesting that you chose to multiply out from the left in the first step as opposed to the right (using (a)). But it's commutative so it really doesn't matter (or maybe I'm simply not making sense anymore). $\endgroup$ – interrogative Dec 22 '15 at 23:36
  • $\begingroup$ @interrogative You're right, I wasn't paying attention and inadvertently skipped the "switch around, then distribute, then switch back" mechanic because, as you say, things are commutative. $\endgroup$ – Alex Provost Dec 23 '15 at 17:03

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