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B is defined as a boolean function on a complex number, $B : \mathbb C \to \mathbb B$, so that B bisects the complex numbers into two regions such that $B(z_1 z_2) = B(z_1) B(z_2) $. What are the possible solutions for $B$?

One such solution is $B(z) = \begin{cases} 0, {if} |z| = 0 \\ 1, {otherwise}\end{cases}$, so that if either $z_1$ or $z_2$ has a magnitude of 0, the product $z_1 z_2$ will have a magnitude of 0, and the condition for $B$ will be met.

Are there any other solutions of $B$? If not, is it provable that $B$ can only have this one solution above?

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  • $\begingroup$ Yes. $B(0) = 17,$ $B(x) = 1,$ whenever $x\neq 0.$ $\endgroup$
    – Igor Rivin
    Commented Sep 13, 2016 at 20:23
  • $\begingroup$ I don't understand your question. What is $B$? Is it a function that takes as input complex numbers? If so, what does it return as output? $\endgroup$
    – Qiaochu Yuan
    Commented Sep 13, 2016 at 20:27
  • $\begingroup$ @Qiaochu Yuan - Yes, B is a function that takes a complex number "x" as input, and returns a number, presumably a 0 or 1, to identify which half of the bisected complex space "x" falls in. $\endgroup$
    – Dave Lampert
    Commented Sep 13, 2016 at 21:24
  • $\begingroup$ @Igor Rivin - if B(0) = 17, then B(0*0) = B(0) * B(0) means that 17 = 17 * 17, which of course isn't correct. That's why B would most likely have values of 0 or 1, but I can't prove that. $\endgroup$
    – Dave Lampert
    Commented Sep 13, 2016 at 21:26
  • $\begingroup$ @DaveLampert Duh, you are right, but see my answer. $\endgroup$
    – Igor Rivin
    Commented Sep 13, 2016 at 21:57

2 Answers 2

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If the partition is $0$ and the rest, there are plenty of such "bisection". Otherwise, if $B$ is non-zero,$B$ restricted to $C^*$ is a multiplicative homomorphism from $C^*$ to $C^*$ whose image is of order $2.$ The kernel has index $2,$ but that's impossible (see https://math.stackexchange.com/questions/1706207/subgroup-of-c-nonzero-complex-with-finite-index). So, $B(x) = 0,$ for some $x \in C^*.$ Your partition is, then $\{0, x\}$ and everything else, on which $B=1.$

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  • $\begingroup$ Thanks for your answer, although to be honest, it's a little over my head. Were you saying that there are other solutions for $B$, or just this one? BTW, I revised by original question to be clearer and to use the TeX notation. $\endgroup$
    – Dave Lampert
    Commented Sep 14, 2016 at 3:13
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You seem to want a function $B:\mathbb{C}\rightarrow\{0,1\}$ which is multiplicative. The all $0$ and all $1$ functions work. The one other option (so the only one which takes on both values) is the one you mentioned: $B(0)=0$ and otherwise $B(z)=1.$

From $B(0)=B(0)B(z)$ we see that $B(0)=1$ forces $B(z)=1$ for all $z.$

Suppose now that $B(0)=0.$ If there is some $x$ with $B(x)=1$ then, for any $z\neq 0,$ we have $B(z)=1$ as $B(x)=B(z)B(x/z).$

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