$2n$ are even integer numbers and $2n+1$ are odd integer numbers. But what are $2n+i$ integer numbers called ($i$ is the imaginary unit)? Are these complex odd numbers? Is there anything interesting to say about these numbers?
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Related (if not duplicate:) [Do odd imaginary numbers exist?](https://math.stackexchange.com/q/26837/42969). – Martin R Aug 19 '19 at 21:08
4 Answers
Parity arithmetic extends to any ring which has $\,\Bbb Z/2 = $ integers $\!\bmod 2\,$ as a homomorphic image, e.g. the Gaussian integers $\,\mathbb Z[i],\,$ where the image $\ \mathbb Z[i]/(i\!\!1,2)\, \cong\, \mathbb Z/2.\, $ In this quotient ring:
$\!\bmod (\color{#0a0}{i\!1},\,\color{#c00}2)\!:\ \ \color{#0a0}{i\equiv 1},\,\color{#c00}{2\equiv 0},\ $ so $\,\ a\!+\!b\,\color{#0a0}i\,\equiv\, a\!+\!b\bmod{\color{#c00} 2}\ \ $ [Gaussian integer parity]
This implies that $\,a\!+\!b\,i\,$ is even $\,\iff 2\mid a\!+\!b\iff a,b\,$ have equal parity.
Similar ideas work for some other rings of algebraic integers (but parity needn't [uniquely] exist in general). See this answer for further discussion. See also this answer which uses parity in $\,\mathbb Z[\sqrt{5}]\,$ to show the integer $\,(9+4\sqrt{5})^n + (94\sqrt{5})^n\,$ is even, and see this answer for parity in an ordered ring with infinite elements (polynomial germs at $\infty$)
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In the Gaussian integers, it makes sense to call multiples of $1+i$ as even, and nonmultiples as odd. The parity of a Gaussian integer $a+bi$ is the same as the parity of $a+b$. In particular, $2n$ is even and $2n+1$ is odd, just as usual; also, $2n+i$ is odd, as you expected!
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The set of complex numbers $\{2n+i:n\in\mathbb{Z}\}$ is a subset of the Gaussian integers, $\{2n+m:n,m\in\mathbb{Z}\}$. As far as I'm aware, there isn't a specific name for it.
As for special properties about them, any special property about the integers will apply to $\frac{zi}{2}$.
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What we mean when we say that an integer $a\in\mathbb Z$ is even is that it belongs to the set $$2\mathbb Z=\{\dots,4,2,0,2,4,\dots\},$$ and what we mean when we say that it is odd is that it belongs to the "shifted copy" of this set $$2\mathbb Z+1=\{\dots,3,1,1,3,5,\dots\}.$$ This gives us the concept of modular arithmetic, which really just means that we identify all the elements of $2\mathbb Z$ with $0$ and identify all the elements of $2\mathbb Z+1$ with $1$, and do arithmetic as usual. This is what gives us the "mod $2$" arithmetic that determines a number's parity.
To do the same thing modulo any integer $m\in\mathbb Z$, what we need to do is to identify all the elements of $m\mathbb Z$ with $0$, identify the elements of $m\mathbb Z+1$ with $1$, and so on, identifying $m\mathbb Z+n$ with $n$ for each $n=0,1,2,\dots,m1$. We can extend this idea to the ring of Gaussian integers $$ \mathbb Z[i]:=\{a+bi\mid a,b\in\mathbb Z\}. $$ First we consider the set $2\mathbb Z[i]$ and identify it with $0$, and we also identify $2\mathbb Z[i]+1$ with $1$. So far so good, but this unfortunately does not exhaust all of the Gaussian integers. For example, $1+i\in\mathbb Z[i]$ does not belong to either of the two categories, whereas any integer belongs to either $2\mathbb Z$ or $2\mathbb Z+1$. To get a complete modular system, we have to cover all the numbers in the ring we're working in, so the resolution is to add two more categories: $2\mathbb Z[i]+i$ and $2\mathbb Z[i]+(1+i)$, and identify them with $i$ and $1+i$ respectively.
We can, of course, now say that $2\mathbb Z[i]$ is the set of "even" numbers, and anything belonging to the latter two categories is "odd". This fits with the definition that even numbers are exactly those that are multiples of $2$, and odd numbers are those that are not multiples of $2$. However, another way to do things is to break up the set $$\{ 2\mathbb Z[i],2\mathbb Z[i]+1,2\mathbb Z[i]+i,2\mathbb Z[i]+(1+i) \}$$ even further, by looking at things modulo $(1i)$. This might seem weird to you, since the modulus is now not an ordinary integer, but a complex number! It is in fact perfectly fine, since inside $\mathbb Z[i]$ they are no different and the ordinary integers are not given any special treatment. What modulo $(1i)$ means is that the set $2\mathbb Z[i]+i$ is identified with $1$ just like $2\mathbb Z[i]+1$ is, and that $2\mathbb Z[i]+(1+i)$ is identified with $0$, just like $2\mathbb Z[i]$. Now, we only have two categories: $(2\mathbb Z[i]+1)\cup(2\mathbb Z[i]+i)$, which is identified with $1$, and $(2\mathbb Z[i])\cup(2\mathbb Z[i]+(1+i))$, which is identified with $0$.
Now, we call the numbers in the latter category even, and that in the former category odd. Done!
For a ring $R$ in general, the set $aR$ of all multiples of $a\in R$ is more commonly known as the ideal generated by $a$, and the idea of looking at "shifted copies" $aR+b$ and identifying them with $b$ is known as "modding out by the ideal $aR$". This is something that is studied in any good beginner's course on rings and fields, should you be interested.
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