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 $\lfloor i\rfloor$?

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    $\begingroup$ No. There is no way to even order the complex numbers in a way consistent with multiplication, so saying $n<i$ is meaningless. $\endgroup$ – Christopher Carl Heckman Apr 30 '16 at 1:27
  • $\begingroup$ 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) $\endgroup$ – Milo Brandt Apr 30 '16 at 1:27
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    $\begingroup$ For interesting stuff like this, Wolfram Alpha can be of big help sometimes. $\endgroup$ – Noble Mushtak Apr 30 '16 at 1:30
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    $\begingroup$ 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 $\endgroup$ – 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 $\Bbb{C}$.


The floor function was intended to be used in the real numbers only.

However, it has been generalized to complex numbers, to be applied on the real part and the imaginary part separately.

For example, generalized floor($0.9+1.1i$) would equal $0+1i$.

Then, the floor of $i$ would be $i$.


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