Quote by D H
There are four solutions if A and B are restricted to the reals. If A and B can be complex numbers or vectors in R^{N}, N>1, the number of solutions is infinite.

Agreed. But we haven't found
all of the complexvalued solutions. Jimmy's solution did not include skeptic2's or my answers.
Spoiler
Geometric argument:
In the complex plane, the numbers 0, A, and A+B form the vertices of a triangle. The vertex at A+B must be a distance √2 from the origin, while the two shorter sides' lengths must add up to 2.
For a fixed value of A+B, say √2 + 0i, the possible values of A forms an ellipse with foci at 0 and A+B, and major axis of length 2.
Furthermore, any solution pair can be multiplied by e^{iθ} to obtain all possible solutions.
Uh, and θ must be real.