Nonstandard Complex Numbers There is an aspect of Non-Standard Analysis that for quite some time I cannot get my head around. I have done some research on this platform to avoid an exact double answer, or one that could be derived, but I still find my question unanswered. I'm happy to receive links to the threads I missed.
So, one can construct Non-Standard reals, say via an ultrafilter. Then, infinitesimals in the nonstandard realm are the equivalence classes of null sequences. Infinite nonstandard reals are equivalence classes of real sequences that are unbounded (positive and negative sign) For alternating sequences like (-1, +1, -1, +1,...), there is an infinite set in the ultrafilter that does determine which of the +1, -1 it is. So all hyperreals are (representatives of equivalence classes of) sequences.
Now, as so many other questions ask, what happens with $\mathbb{C}$?
It is clear that the hyperreals are elementary equivalent to the reals, and categoricity for the complex numbers make them isomorphic to any completion of the hyperreals in that same sense. So what happens to all the infinitesimals? Do the hyperreal non-zero infinitesimals suddenly become hyperreal complex null-sequences? While when going the 'standard way' they should be zero all along? In particular, what is the function $f$ in the following diagram:
$\require{AMScd}$
\begin{CD}
    \mathbb{R} @>N>> \mathbb{R}^*\\
    @Vc  V V @VV  CV\\
    \mathbb{C} @>>f> \mathbb{C}\equiv\mathbb{C}^{*}
\end{CD}
Here, $N$ is the usual ultrapower embedding, $c$ and $C$ are the respective algebraic completions.
The particular question is, what happens to $(1/n)_{n\in\mathbb{N}}$? That is a non-zero infinitesimal, equivalent to zero taking standard parts, so it should be a zero non-standard complex number, but cannot be going the non-standard complex way. What is wrong with this? Or rather, what implicit projection are we missing? And for the set theorists, what kind of cardinal do we need to exist for this to work?
 A: The story for $\mathbb{C}$ is exactly like the story for $\mathbb{R}$, and nothing weird happens like you are asking about.  We have elements of $\mathbb{C}^*$ which are "infinite" or "infinitesimal".  A nonzero infinitesimal hyperreal is still a nonzero infinitesimal when considered as an element of $\mathbb{C}^*$.
Your point of confusion seems to stem from the fact that $\mathbb{C}$ and $\mathbb{C}^*$ are isomorphic as fields.  This is true, but is pretty much irrelevant.  Yes, there exists an isomorphism between them.  So what?  There is no canonical choice of such an isomorphism, and such an isomorphism is just an isomorphism of fields and does not respect any other structure about $\mathbb{C}$ which we might care about (such as the subfields $\mathbb{R}$ and $\mathbb{R}^*$).
It is true that you can choose some field isomorphism $g:\mathbb{C}^*\to\mathbb{C}$, and then ask what happens to $\mathbb{R}^*$ (or to the standard copy of $\mathbb{C}$ inside $\mathbb{C}^*$) under this isomorphism.  The answer is that you can't really say much, since this $g$ is highly non-unique.  Indeed, it is possible to choose $g$ to map any transcendental element of $\mathbb{C}^*$ to any transcendental element of $\mathbb{C}$ (where "transcendental" means "not algebraic over $\mathbb{Q}$").  So, for instance, the element of $\mathbb{C}^*$ represented by the sequence $(1/n)$ could map to any transcendental number in $\mathbb{C}$ at all, depending on the choice of isomorphism $g$.
