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 http://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 $B=1.$

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 $B=1.$

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. So, $B(x) = 0,$ for some $x \in C^*.$ Your partition is, then $\{0, x\}$ and everything else, on which $B=1.$

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 http://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 $B=1.$