# What is $\lfloor i\rfloor$? [duplicate]

So, floor is a function that converts a real number to an integer. It rounds down. This makes sense; however, what about complex numbers? I know that depending on the number, it can be split linearly. However, I do not know what $\lfloor i\rfloor$ equals. Is there a clearly defined definition of $\lfloor i\rfloor$?

• No. There is no way to even order the complex numbers in a way consistent with multiplication, so saying $n<i$ is meaningless. – Christopher Carl Heckman Apr 30 '16 at 1:27
• It depends what you want it to do; floor basically tells us which interval of the form $[n,n+1)$ a particular number is. On some sense, this is intuition based on the order - so doesn't extend to $\mathbb C$, which can't be ordered. However, if you're interested in what rectangle of the form $[n,n+1)+i[m,m+1)$ a number is in, you just floor each coordinate (but this the lattice $\mathbb Z + i\mathbb Z$ in $\mathbb C$ doesn't bear the same importance as $\mathbb Z$ does in $\mathbb R$, so the definition isn't as natural) – Milo Brandt Apr 30 '16 at 1:27
• For interesting stuff like this, Wolfram Alpha can be of big help sometimes. – Noble Mushtak Apr 30 '16 at 1:30
• note that you can define some sort of pseudo-analytic continuation of $\lfloor x \rfloor = x - \{x\} = x - 1/2 + \sum_{n = 1}^\infty \frac{\sin(2 \pi n x)}{\pi n}$ with $\displaystyle f(z) = z + \frac{\ln(1-e^{2 i\pi z}) - \ln(1-e^{-2 i\pi z})}{2 i \pi}$ in some cases it can work, it even implies the functional equation for the Riemann zeta function – reuns Apr 30 '16 at 1:45

$\Bbb{C}$ is not totally ordered, and thus the floor function can't really be defined conventionally on it.
You could think of the floor of $z$ as $\lfloor \text{Re} \ z\rfloor+i*\lfloor \text{Im} \ z\rfloor$, which in the case of $z=i$, would be $i$, but in general, the floor is not defined on $\Bbb{C}$.
For example, generalized floor($0.9+1.1i$) would equal $0+1i$.
Then, the floor of $i$ would be $i$.