Do odd imaginary numbers exist?

"Odd" has several meanings in mathematics: you have odd integers (those which are not multiples of $2$); you have odd functions (those that satisfy $f(-x) = -f(x)$ for all $x$); and possibly others.

If you want to stick to the first meaning, then there's two things to keep in mind: even for just real numbers, "odd" in the sense of "not a multiple of $2$" doesn't really work very well, because every real number is a multiple of $2$: given $r\in\mathbb{R}$, $r = 2\left(\frac{r}{2}\right)$, and $\frac{r}{2}$ is a real number. The same is true for complex numbers: if $a+bi\in\mathbb{C}$, then $a+bi = 2\left(\frac{a}{2} + \frac{b}{2}i\right)$, so every complex number would be "a multiple of $2$", so no complex number would be odd. So this concept does not really do much for complex numbers as a whole.

On the other hand, you can restrict to those complex numbers which have integer real and complex part: $a+bi$ with $a,b\in\mathbb{Z}$ (instead of $a,b\in\mathbb{R}$, like in $\mathbb{C}$). These are called the Gaussian integers because they were first studied by Gauss.

For these numbers, you can talk about "multiples of $2$": a Gaussian integer $a+bi$ is a multiple of $2$ if and only if both $a$ and $b$ are even: because if $a+bi = 2(x+yi)$ with $a,b,x,y\in\mathbb{Z}$, then $a=2x$ is even and $b=2y$ is also even. So the "odd Gaussian integers" would be all Gaussian integers that are not multiples of $2$, namely the $a+bi$ that have either $a$ or $b$ odd. Note that $1$ would be an "odd Gaussian integer" (which looks good, because $1$ is an odd integer), but then again so would $1+2i$ (which may not look so good).

There is also a slight wrinkle: in the integers, if you add two integers of the same parity, you will always get something that is "even", and if you add two integers of different parity you will get something that is "odd." This does not happen with the above notion of "even" in the Gaussian integers. For instance, $1+2i$ is "odd", and so is $2+i$; if we add them, we get $3+3i$ which is also "odd." In fact, we have four different kinds of Gaussian integers: the "even" ones (both real and imaginary parts are even); the "even-odd" ones (real part even, imaginary part odd); the "odd-even" ones (real part odd, imaginary part even); and the "odd" ones (both real and imaginary part odd). It is only if you add two of the same kind that you will get an "even" Gaussian, and if you add two different kinds you will get an "odd" Gaussian. So this concept of "even" and "odd" does not seem to behave like it does in the integers. Added. What is worse, as Bill points out in comments, even this does not work well with multiplication, since for example the product of an "even-odd" by an "even-odd" gives an "odd-even", not an "even-odd".

We might, then, want to look at another possibility.

Another possibility is to notice that $2i = (1+i)^2$, and $i$ is invertible in the Gaussian integers. So instead of looking at the multiples of $2$, you can try looking at the multiples of $1+i$ (just like you don't define "odd" in terms of multiples of $4$ in the integers; I bring up $4$ because $4=2^2$). When is a Gaussian integer a multiple of $1+i$? $$(x+yi)(1+i) = (x-y) + (x+y)i.$$ Can we recognize such numbers? I claim they are precisely those Gaussian integers $a+bi$ with $a+b$ even.

Indeed, if $a+bi$ is a multiple of $1+i$, then as above we have $a=x-y$ and $b=x+y$ for some integers $x$ and $y$, so $a+b = (x-y)+(x+y) = 2x$ is even. Conversely, suppose that $a+bi$ has $a+b$ even, $a+b = 2k$. Then $a-b$ is also even (since $a-b = (a+b)-2b$), so we can write $a-b = 2\ell$. Then \begin{align*} (k -\ell i)(1+i) &= (k+\ell) + (k-\ell)i\\ &= \left( \frac{a+b}{2} + \frac{a-b}{2}\right) + \left(\frac{a+b}{2} - \frac{a-b}{2}\right)i\\ &= a + bi, \end{align*} so $a+bi$ is a multiple of $1+i$. So if you define "odd" in terms of "multiple of $1+i$, then it corresponds precisely to whether or not $a\equiv b\pmod{2}$: if $a$ and $b$ have the same parity, then $a+bi$ is "even"; if $a$ and $b$ have different parity then $a+bi$ is "odd".

It also has the advantage of mirroring a bit better what happens with parity in the integers: if you add two "even" or two "odd" Gaussian integers (under this definition), then the sum is "even"; and if you add an "even" and an "odd" Gaussian integer you get an "odd" Gaussian integer. Also, if you multiply an "even" Gaussian by any Gaussian you get an "even" Gaussian: for if $a$ and $b$ have the same parity, then $$(a+bi) (x+yi) = (ax-by) + (ay+bx)i.$$ If both $a$ and $b$ are even, then so are $ax-by$ and $ay+bx$, so the result is even. If both $a$ and $b$ are odd, then either $x$ and $y$ have the same parity, in which case both $ax-by$ and $ay+bx$ are even; or else $x$ and $y$ have different parity, so that both $ax-by$ and $ay+bx$ are odd. Either way, the product is "even." Similarly, if you multiply two "odd" Gaussians, the result will be "odd."

So I think the latter concept is a bit more intuitive, but that may be just me.

Post data. There is in fact a lot of very interesting stuff in the background of the above; considering $1+i$ instead of $2$ in the Gaussian integers comes from Algebraic Number Theory.

As I mentioned in an earlier post on casting out nines, one can apply parity arguments in any ring that has $\, \mathbb Z/2\, $ as an image, e.g. all rationals with odd denominator, or the Gaussian integers $\,\mathbb Z[i],\,$ where the image $\ \mathbb Z[i]/(2,i-1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\ a+b\,i\ $ is even iff $\ a\equiv b\pmod{\!2}.\,$ i.e. if $\ a+b\,i\ $ maps to $\,0\,$ via the above isomorphism, which maps $\ 2\to 0,\ i\to 1$.

Generally it's easy to show that if $\,2\nmid f(x)\not\in\mathbb Z\,$ then the number of ways to define parity in the ring $\ \mathbb Z[w] \cong \mathbb Z[x]/f(x)\ $ is given by the number of roots of $\, f(x)\, $ modulo $2.\, $ For suppose that there exists a homomorphism $\ h\, :\, \mathbb Z[w]\to \mathbb Z/2.\,$ Then $\,w\,$ must map to a root of $\,f(x)\,$ in $\ \mathbb Z/2.\, $ Thus if $\ f(0)\equiv 0\pmod{\!2}\ $ then $\, \mathbb Z[w]/(2,w) \cong \mathbb Z[x]/(2,x,f(x)) \cong \mathbb Z/2\ $ since $\, x\mid f(x)\pmod{\!2}.\, $ Similarly if $\ f(1)\equiv 0\pmod{\!2}\ $ then $\, \mathbb Z[w]/(2,w\!-\!1) \cong \mathbb Z/2\ $ by $\, x\!-\!1\mid f(x)\pmod{\!2}. $

Let's consider some simple examples. Note $\ x^2+1\ $ has the unique root $\ x\equiv 1\pmod{\!2},\,$ so the Gaussian integers $\ \mathbb Z[i]\cong \mathbb Z[x]/(x^2+1)\ $ have a unique definition of parity - with $\,i\,$ being odd. Since $\ x^2+x+1\ $ has no roots modulo $\, 2,\, $ there is no way to define parity for the Eisenstein integers $\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2+x+1).\, $ Indeed since $\ w^3 = 1\ $ we infer that $\, w \equiv 1\pmod{\!2}\, $ contra $\ w^2+w+1 = 0.\, $ On the other hand $\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2+x+2)\ $ has two parity structures since both $\,0\,$ and $\,1\,$ are roots of $\ x^2 + x + 2\ $ modul0 $\,2,\,$ so we can define $\,w\,$ to be either even or odd. A simple application is in this post where we use parity in the ring $\,\mathbb Z[\sqrt{5}]\,$ to prove that the integer $\,(9+4\sqrt{5})^n + (9-4\sqrt{5})^n\,$ is even.

Any parity argument that's truly ring-theoretic will generalize to any ring $\,R\,$ with parity, e.g.

Parity Root Test $\ $ Suppose $\,f(x)\,$ is a polynomial with coefficients in a ring $\,R\,$ with parity. Then $\,f(x)\,$ has no roots in $\,R\,$ if $\,f(x)\,$ has constant coefficient and coefficient sum both being odd. Further, $\,f(x)\,$ has no roots in the fraction field of $\,R\,$ if $\,R\,$ is a domain and the leading coefficient of $\,f(x)\,$ is odd, and $\,0\,$ is the only element of $\,R\,$ that is divisible by arbitrarily large powers of $\,2.$

Proof $\ $ The test simply verifies $\ f(0)\equiv f(1)\equiv 1\pmod{\!2}\ $ i.e. $\, f(x)\,$ has no roots $\bmod 2.\, $ Therefore $\,f(x)\,$ has no roots in $\,R.\,$ For the fractional case, by the hypothesis, we may cancel powers from a fraction $\,a/b\,$ until $\,a,b\,$ are not both even. If $\,b\,$ is odd then $\,a/b\equiv a\pmod{\!2}\,$ would be a root, so $\,b\,$ is even and $\,a\,$ is odd. Then $\,0 = b^n\,f(a/b) = f_n a^n + b\,(\cdots) \equiv 1\pmod{\!2},\,$ a contradiction, since the leading coefficient $\,f_n\,$ and $\,a\,$ are both odd and $\,b\,$ is even.

As a corollary, we deduce that no ring $\,R\,$ with parity contains a primitive cube root of one. Indeed, if so, $\,R\,$ contains a root of $\ f(x) = x^2+x+1\,,\ $ contra $\,f(x)\,$ has no roots in $\,R\,$ by the parity root test. Hence, for example, the Gaussian integers do not contain a primitive cube root of one.

Sure. If you define an even number as any number $k$ expressible as $k=2z$ where $z$ is a Gaussian integer, then an odd number would be any number $k$ for which no such $z$ can be found. This would be any number whose real or imaginary part (or both) is odd.