I apologize if 'consistent' is not the correct term here, in any case, I'll try my best to explain my doubt.

Say we have $n$ complex numbers $k_1,k_2,...,k_n$, and we create an equation with them using only addition and multiplication, for example

$$k_1k_2-7k_3+5k_4^2$$, or


etc... (To elevate to a constant integer is allowed since is just repeated multiplication, $k_3^3=k_3k_3k_3$)

Now, say we have a certain equation composed of specific complex numbers, with the operators of addition and multiplication. Such equation should only take one value, but how can we be sure? To be clearer, given an equation as the ones described, how could one be sure the following conditions hold?

Condition 1: no matter the order in which we perform the operations, the equality holds. (Clearly this excludes things like $ab+c=a(b+c)$)

Condition 2: even after substituting $k_i$ with an expression -again, one that uses the same two operations- that is equivalent to it, the value of the equation is the same and condition 1 still holds.

If one is to write an equation using only addition, then is clear that both conditions hold, since complex numbers depend on an imaginary value and a real one, both which won't change even if the numbers in the equation are written differently (using an expression comprised only of addition). For multiplication, noticing that each complex numbers depends on an angle and a magnitude and that it can be written as $ae^{ix}$ helps to easily arrive at the same conclusion.

So both condition one and two would hold if we were to only use one operator. From this, it becomes intuitive that both conditions should still hold even when the equations used are a combination of both addition and multiplication. But, I haven't been able to prove it. How could one do that?

I would really appreciate any help/ideas.

  • $\begingroup$ I am confused by your claim that "both conditions should still hold even when the equations used are a combination of both addition and multiplication". Isn't $ab+c$ a counterexample to that, as you mentioned earlier? $\endgroup$ – Eric Wofsey Dec 16 '17 at 1:21
  • $\begingroup$ Also, your use of the term "equation" is nonstandard and very confusing. An "equation" is a statement that two things are equal. $\endgroup$ – Eric Wofsey Dec 16 '17 at 1:22
  • $\begingroup$ Again, I apologize for my lack of math vocab. An example in which condition 1 wouldn't hold is if $(ab+c)+(d(e^2))≠(ab)+(c+(de)e)$. Though, I think I see now how condition 1 is true. I'm still trying to understand #2. $\endgroup$ – Leo Dec 16 '17 at 1:32
  • $\begingroup$ If $ab+c$ does not give a counterexample to condition 1, I'm afraid I have no idea what condition 1 is supposed to mean. $\endgroup$ – Eric Wofsey Dec 16 '17 at 1:34
  • $\begingroup$ @Leo How, for example, did you prove an analogous result about real numbers? $\endgroup$ – arseniiv Dec 16 '17 at 2:03

If i understand correctly, you are asking if the complex addition and multiplication are well-defined. Complex multiplication for example is a well-defined function $f\colon\mathbb C\times\mathbb C\to\mathbb C$ sending $(a+bi,g+di)$ to $ag-bd+(ad+bg)i$. The fact that $f$ is a function, (you can prove it), means that for $a+bi=q+wi$ and $g+di=r+ti$ you have $f(a+bi,g+di)=f(q+wi,r+ti)$, so the same result will occur if you write differently the terms of the multiplication. Go read some Algebra

  • $\begingroup$ I'm able to see that if $a+bi=q+wi$ then $f(a+bi, k)=f(q+wi,k)$, in fact I tried to explain that in one of the paragraphs of my question (the one that starts with "If one is to write..."). Now, what if we wrote $a+bi$ as $k_1-6k_2^2+7k_3k_4$, how could one prove then that multiplication is still well defined? Also, $f(x,y)$ is simply multiplication, are all functions that can be written using addition and multiplication also well defined? $\endgroup$ – Leo Dec 16 '17 at 1:44
  • $\begingroup$ Well, any expression which is a result of applying functions from some $C^k$ to $C$, starting with several variables and constants from $C$, represents a function from $C^\text{variable count}$ to $C$. Now take $C=\mathbb C$. $\endgroup$ – arseniiv Dec 16 '17 at 2:01

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