What is the value of $\left \lfloor \frac{1}{3} + \frac{i}{2} \right \rfloor $? How do you get the floor function of a complex number? Is it defined? In particular, for example, what is the value of $\left \lfloor \frac{1}{3} + \frac{i}{2} \right \rfloor $?
I do not know the answer to this problem. This is merely out of curiosity, no special reason for posting.
 A: Note that since there is no order amongst the complex numbers, there is no way to find an integer $s =  [z]$  such that, $z - 1 < s \leq z $ , that you require.
A: Interesting idea. It hasn't really been defined for complex numbers, but I can propose a way to define it analogously.
On the number line, if a number $x$ is on the line segment between the two nearest integers, I visualize the floor function as "pushing" $x$ to the left. In the complex plane, a number $z$ might not fall on a line segment, but in the middle of a unit square formed by integer real and imaginary parts. If this is so, we might define the floor function as "pushing" $z$ to the southwest corner of the square. This is equivalent to flooring both the real and imaginary parts of $z$. So we might define $\lfloor z \rfloor$ on complex $z$ by defining
$$\lfloor a+bi\rfloor:=\lfloor a\rfloor+\lfloor b\rfloor i$$
with $a,b\in\mathbb R$.
If this is the case, then we would have
$$\bigg\lfloor \frac{1}{2}+\frac{i}{2}\bigg\rfloor=0$$
A: No, the floor function is not defined for complex numbers. Since the complex numbers have no order, "the greatest integer less than 1/3+i/2" has no meaning.
A: It is not defined, but it can be defined, but you may not like the answer. 
The reason you can't define one straightforwardly is that the complex numbers do not admit an ordering. As much has been said by other responders but I think there's more to say than "no you can't."
First, consider only complex numbers with rational real and imaginary parts.
Since there is a some relationship between an integer quotient and floor, a related problem that can shed some light on the question is to find the GCD of two gaussian integers, numbers of the form $a+bi,~ a,b\in \mathbb{Z}$. The Euclidean algorithm to find GCD using quotients of Gaussian integers (defined in a moment) will terminate but the GCD you calculate is only unique up to multiplication by $1, -1, i,~ \mathrm{and}~ -i$. With the GCD of two integers we can impose uniqueness by just choosing the positive one, but there's no way to make this choice in the complex numbers. The question is really, then, what we would mean by the quotient of two Gaussian integers.
Your question about the floor of $\frac{1}{3} + \frac{1}{2}i$ can be seen as asking for the quotient of $\frac{2+3i}{6+0i}$ and both $0+0i$ and $0+1i$ would be acceptable results. Remember that the quotient and remainder of $\frac{a}{b}$ satisfy $a = bq+r$ and we select $q$ so that $q$ is the largest integer leaving $r\geq 0$. But there is no such unique choice in $\mathbb{C}$ because you cannot order $\mathbb{C}$ by $\geq$ meaning we can select different $q$. In this case if we choose $q=0$ then $r$ is $2+3i$, and if we choose $q=i$ then the remainder is $2-3i$. Both of these have the same magnitude of $\sqrt{13}$ so which is the "right" one to minimize the remainder?
So how to define the quotient of two Gaussian integers? Perform ordinary complex division by rationalizing the denominator and distributing, then round the real part and imaginary part. It doesn't matter if you round $\frac{1}{2}$ to $0$ or $1$, as the previous paragraph shows, but it definitely matters that e.g. $\frac{3}{4}$ rounds to $1$. For the sake of completeness I should mention that the ordinary GCD algorithm also doesn't care if you use rounding, it will still give you the right GCD up to multiplication by a unit (namely $-1$). Unlike the integers, however, if you attempt to only use floor with Gaussian integers you can enter an endless cycle and the algorithm will not terminate.
Given the foregoing, $\lfloor a+bi\rfloor,~ a,b\in \mathbb{R}$ should be the complex number $\lfloor a\rceil + \lfloor b\rceil i$.
