Why does WolframAlpha give a strange solution set to $\lfloor a\rfloor=0$?

Question

I decided to go and look up the roots of $\lfloor a\rfloor$ in WolframAlpha where $a$ is any number, real or complex, and of course the interval $[0,1)$ showed up as an answer but I also got these answers as the complex solutions!

These numbers seem fairly random to me so now I'm curious, what on earth do these numbers represent and how did WolframAlpha come up with them?

EDIT:

Interestingly enough, I decided to type in $\lfloor a+bi\rfloor$ into WolframAlpha and it told me that it was equivalent to saying $\lfloor a\rfloor+i\lfloor b\rfloor$. The weird part about this is that that means that any complex number $a+bi$ where $a$ and $b$ are in the interval $[0,1)$ should cause $\lfloor a+bi\rfloor$ to be equal to $0$. However, the only answers that WolframAlpha give are the ones that I posted above... So what is going on here?!?!?

Also, someone in the comments mentioned that the answers appear to be $\frac{\sqrt{3}}{2} + \frac{1}{2}i$ and $\frac{1}{2} + \frac{\sqrt{3}}{2}i$.

Answer 1

I suppose the answer might look a little silly but note that floor(1/2+sqrt(3)*i/2) returns zero, so the result is indisputably correct.

Typically, WolframAlpha will try to classify the results to some extant. If you just enter x^3-x-1=0, for example, you'll see separate pods containing a real solution and two complex solutions. Now in your case, Mathematica is not able to solve floor(a)=0 over the complexes using analytic tools so it falls back to a numerical search and finds the specific examples you mentioned.

Similar behavior is exhibited when we enter round(x)=i:

Again, Round[-1/2+Sqrt[3]I/2] yields I in Mathematica, so the result is correct. Perhaps a bit silly, but not totally unreasonable either.