Are the imaginary zero, the complex zero, and the real zero distinct numbers? The additive identity element is unique. Does this imply that all zeros are not distinct?
Am struggling to explain how a unique additive element does not rule out more than one number zero. I am thinking it only rules out using more than one zero at a time in an arithmetic that includes addition.
 A: "$0$" is just a name given to a particular element of some sets (in this case, an abelian group.)
In two disjoint sets, each one can have a member playing the role of $0$, and since the sets are disjoint the two zeros have to be distinct.
For a subgroup of a group, though, their identities must be the same element. In that case, there is definitely only one element playing the role of the identity in both sets.
Most of the time we think of the integers, rationals, reals and complex numbers as being a chain in which each is a subgroup (a subring even) of the next, and in that picture they all share the same additive identity (and multiplicative identity, for that matter.)
But in principle one could specify two disjoint sets, both isomorphic to $\mathbb Z$, and it would be the case that there are "two zeros," one in each copy.
I could be getting the wrong impression, but the question posed seems to imply that there is some sort of "absolute zero" that must be shared between all additively written abelian groups, which is not the case.  "$0$" is just a special notation for a particular element in a set. It's not a universal constant.  Apologies if the impression I am laboring under is incorrect.


@rschwieb "Imaginary zero" vs "complex zero" is 0i and 0+0i. The imaginaries are numbers in their own right

This represents a misapprehension about notation, I think. Writing "$0+0i$" is a little ambiguous. When representing complex numbers as pairs of real numbers, each complex number is uniquely represented as $(a,b)$ for two real numbers $a,b\in\mathbb R$.
Now, one can modify the notation by introducing the following conventions:
$$i=(0_\mathbb R,1_\mathbb R)$$
$$1_\mathbb C=(1_\mathbb R,0_\mathbb R)$$
$$0_\mathbb C=(0_\mathbb R,0_\mathbb R)$$
, and then one can correctly write that $a1_\mathbb C+bi=a(1_\mathbb R,0_\mathbb R)+b(0_\mathbb R,1_\mathbb R)=(a,b)$.
As you can see,  $0_\mathbb R1_\mathbb R+0_\mathbb R i=(0_\mathbb R,0_\mathbb R)=0_\mathbb R i$ are the same element, so there's no point in distinguishing them.
In this scheme, $0_\mathbb R$, viewed only as a coefficient of a linear combination in $\mathbb C$, is not the same thing as $0_\mathbb C$.  To view $\mathbb R$ as a subset of this model of $\mathbb C$, one has to make the further identification $a\mapsto a1_\mathbb C=(a,0)$, which of course identifies $1_\mathbb R$ with $1_\mathbb C$.
