Almost four years and no answers? I can't give you a complete answer, but I guess under the theory of better late than never I will try to address as many as your points as I can. Though it's possible that by now you have gotten a better answer through another source.
Does anyone have any good resources on how complex number systems
work?
Yeah. Go to Wikipedia and scroll all the way to the bottom, to find the external links. The Gilbert paper looks very good and transparent, and it's available as a PDF. Your local university library might have a copy of Knuth's monumental 3-volume book.
there are a couple of things I don't quite understand (why there are 4
digits, for example).
This is a very interesting question in its own right. I think that first you need to understand "negabases" (numeral systems with a negative number as the base), but before that, I will review some very basic basics of positive integer bases.
You already know, at least on an intuitive level, the basic principle of power series numeral systems with a positive integer as the base. Given a base $b$ and a positive integer $n$ represented by the string of $\mathcal{L}$ digits $\{d_{\mathcal{L} - 1}, d_{\mathcal{L} - 2}, \ldots d_1, d_0 \}$, the string of digits is essentially a shorthand for $$n = \sum_{i = 0}^{\mathcal{L} - 1} d_ib^i$$. Of course this basic principle also works when $b$ is almost anything other than a positive integer greater than 1.
I could be wrong on this next point, but I think that there are only three numbers that are completely unworkable as bases for a power series positional numeral system, namely $-1$, 0 and 1. I think that for a base $b$ to be "workable" for this purpose, the following needs to be true: if $\alpha$ and $\beta$ are distinct integers drawn from $\mathbb{Z}$, then $b^\alpha$ and $b^\beta$ are distinct numbers.
With $b = -1$, $\alpha$ and $\beta$ can be distinct but if they're of the same parity, then $b^\alpha = b^\beta$, and with $b = 1$ they can have different parities. It should be obvious that 0 is completely useless as a base. Other than that, I guess any number whatsoever can work as a base. Of course some bases have more practical value than others. For example, quite a few people think we'd all be better off replacing 10 with 12 as our main number base, but I've never heard anyone say the same for $\sqrt[3]{1729}$.
With complete confidence I can tell you that there are only two digits you must absolutely have in every power series positional numeral system: 0 and 1. You have to have 0 to signify the absence of a specific power of the base. Whether you need any more digits depends on the choice of base. Given an integer $b > 2$, you also need a digit for $b - 1$; in fact you need to be able to represent the integers from 0 to $b - 1$ with a single digit.
It is this need for digits from 2 to $b - 1$ when $b > 2$ that leads to some frustrated expectations when the base is not a positive integer. For example, for $b = 4$, you need digits for 0, 1, 2, 3. You don't need a digit for 4 because that's represented as 10. What digits do you need for "negaquartal," with $b = -4$? Digits for $-1$, $-2$, $-3$? Actually, no (though I suppose that if I really wanted to, I could try to work out a system using such digits; I have not wanted to).
In a philosophical sense, systems with positive bases are incapable of representing negative numbers. Sure, we have the negation operator, but in a sense that is actually a shorthand for "$0 -$." Which is to say that, e.g., $-7$ doesn't represent negative seven, that it actually represents $0 - 7$. It is this perceived shortcoming that negabases address: they can represent negative numbers without the need for a negation operator.
For negaquartal, you need digits for 0, 1, 2, 3. The numbers $-1$, $-2$, $-3$ are not single-digit numbers in this system. But you can't use 10 for 4 because now that means $-4$. So positive 4 is represented as 130, and in general, for $\alpha$ odd, the negaquartal representation of $4^\alpha$ is 13 followed by $\alpha$ 0s; this is tantamount to $4^{\alpha + 1} - 3(4^\alpha)$. The negaquartal representation of $4^\alpha$ with $\alpha$ even is the same as with positive quartal. As for $-3$, that's 11, which means $-4 + 1$.
To represent complex numbers, both positive bases and negative bases have to separate the real part from the imaginary part. That's where complex number bases come in. Quater-imaginary, as you know, has $b = 2i$. As you may have already realied, $2i$ is the principal square root of $-4$. So the way it works out is that to obtain the quater-imaginary representation of a purely real integer, be it negative or positive, you can just take the negaquartal representation and riffle in some zeroes. Thus, $-3$ is 11 in negaquartal and 101 in quater-imaginary. I know I've been longwinded on this one without fully answering your question, but if you've read this far and you understand why quater-imaginary uses 4 digits, I'm happy.
An unfortunate truth about converting a number from base $a$ to base $b$
is that unless there are integers $m > 0$ and $n > 0$ such that $a^m = b^n$,
you need to use an arbitrarily large amount of memory in order
to convert arbitrarily large numbers from base $a$ to base $b$ -- not necessarily all the digits all the time, but a lot of the digits a lot of the time.
When $a^m = b^n$ (such as the conversion between $a=2^8$ and $b=2^5,$ which was mentioned in the question, where $a^5 = b^8$), you are able to "chunk" the input string of digits into groups of $m$ digits that can be converted into groups of $n$ digits, one group at a time with no kind of "carry" or interaction between the groups other than simply concatenating the digits to the output, exactly as described in the question; but that is a special case.
Suppose there are no such numbers $m$ and $n$.
There are various ways this can happen, classified by the prime
factorizations of $a$ and $b$:
- At least one of the bases $a$ or $b$ has a prime factor the other does not have, for example base $8 = 2^3$ and base $10 = 2\cdot5$.
- The bases $a$ and $b$ have the same prime factors, but not in the same ratio of powers, for example base $10 = 2\cdot5$ and base $20 = 2^2\cdot5$,
or base $60 = 2^2\cdot3\cdot5$ and base $90 = 2\cdot3^2\cdot5$.
Note that in the case where $a$ and $b$ are both powers of the same prime,
there is always a way to read and convert one block of digits at a time, provided that you either start with the least significant digit or you know how many digits each number contains.
In the case where $b$ has a prime factor that $a$ does not have,
there is no power of $a$ that will ever be divisible by $b$,
so if we take an arbitrarily large $m$, we can make $a^m$ have as
many significant digits in base $b$ as we want.
So when converting an arbitrarily large number from base $a$ to base $b$,
there is no limit on the size of the block of digits of the output
that could be affected by whether some high-place-value digit of input
happens to be $0$ or $1$.
Conversely, when converting from base $b$ to base $a$,
there is no limit on the size of the block of digits of the input
that we would have to examine in order to know for sure
what the value of some high-place-value digit of the output is.
So both conversion directions take unbounded memory if one base
has a factor the other does not have.
In the case where the two bases have the same prime factors, but the
powers of the prime factors are not in the same ratio, for any
large-enough integer $m$, $a^m$ will be divisible by $b^n$ for some
integer $n > 0$. But the quotient $a^m/b^n$will always contain
some number of "uncancelled" prime factors, and we can show that the
number of these factors grows without bound as $m$ grows arbitrarily large.
At this point I want to introduce some mathematical notation:
for $p$ a prime, $\nu_p(x)$ is the number of factors of $p$
in a prime factorization of $x$; for example, $\nu_2(60) = 2$
because $60 = 2^2\cdot3\cdot5$.
Now for each prime factor $p$ of $a$ and $b$, evaluate $\nu_p(a)/\nu_p(b)$;
let $q$ be the prime factor of $a$ and $b$ that minimizes this ratio, $r$ the prime factor that maximizes it.
Then $\nu_q(a^{\nu_q(b)}) = \nu_q(b^{\nu_q(a)})$; in fact, $a^{\nu_q(b)}$
is divisible by $b^{\nu_q(a)}$, because for every other prime factor $p$,
$\nu_p(a^{\nu_q(b)}) \geq \nu_p(b^{\nu_q(a)})$.
But the quotient $a^{\nu_q(b)}/b^{\nu_q(a)}$ is not divisible by $b$;
it has at least one prime factor of $r$ and no prime factor $q$.
That is, $a^{\nu_q(b)}$ has at least one "uncancelled" factor of $r$
when we take the greatest power of $b$ that divides $a^{\nu_q(b)}$.
If we next consider $a^{2\nu_q(b)}$,
the highest power of $b$ that divides it is $b^{2\nu_q(a)}$,
and after that division there are at least two "uncancelled"
factors of $r$.
In general, $a^{k\nu_q(b)}$ has at least $k$ "uncancelled" factors of $r$;
to write $a^{k\nu_q(b)}$ in base $b$, we would have to write some multiple
of $r^k$ not divisible by $b$ followed by some zeros.
This base-$b$ numeral could have as many significant digits as we want,
depending on how large $k$ is.
So once again, an arbitrarily large number of digits of the base-$b$
output can depend on the value of a single high-place-value digit of
the base-$a$ input, and there is no bound on how much memory we might
require to convert an arbitrarily large number.
The conversions where we can convert blocks of digits one at a time in
bounded memory are just the ones where the two bases have exactly the same
prime factors, and the powers of those prime factors are in the same proportions. This implies both bases are powers of the same integer.
For example, every three digits in base $36 = 2^2\cdot3^2 = 6^2$
can be converted to two digits of base $216 = 2^3\cdot3^3 = 6^3$.
Note that base-85 encoding of a string of (for example)
$1024$ binary digits does not produce a number written in base $85$.
Instead, it produces something like a mixed-radix number, but
with more complex rules for "carryover" than more well-known
mixed-radix systems (such as the time of day) typically have.
Reading from right to left, the place values of a base-85 encoding
of a $1024$-bit integer would be $1$, $85$, $85^2$, $85^3$, $85^4$,
$2^{32}$, $85\cdot2^{32}$, $85^2\cdot2^{32}$,
$85^3\cdot2^{32}$, $85^4\cdot2^{32}$,
$2^{64}$, $85\cdot2^{64}$, and so forth.
You might say that the so-called "base-85" encoding really encodes
a string of bits as a base-$2^{32}$ numeral, except that rather than
define a sequence of $2^{32}$ unique glyphs for the digits
$0$ to $2^{32}-1$, we write each base-$2^{32}$ digit as a
five-digit base-$85$ integer in that range, left-padded with zeros as needed.
By the way, for a conversion from base $a$ to base $b$ where the bases are not powers of the same integer and therefore there is no simple $n$-digit-to-$m$-digit conversion, it is still not ever necessary to use a division algorithm. You just have to work out how to multiple a base-$b$ number by $a$ in base $b$. For example, $723_{10}$ to base $16$:
First represent base $a=10$ in base $b=16$. In base $16$, the number $10_{10}$ is written $\mathrm a_{16}$.
Take the leftmost digit, $7.$ The intermediate result is $7_{16}.$
Multiply by $\mathrm a_{16}$. Result:
$7_{16} \times \mathrm a_{16} = 46_{16}.$
Add the next digit from the input, $2.$ Result: $48_{16}.$
Multiply by $\mathrm a_{16}$. Result:
$40_{16} \times\mathrm a_{16} + 8_{16} \times\mathrm a_{16}
= 280_{16} + 50_{16} = 2\mathrm{d}0_{16}.$
Add the next digit from the input, $3.$ Result: $2\mathrm{d}3_{16}.$
Since that is the last digit of input, the final result is $2\mathrm{d}3_{16}.$
When base $b$ is smaller than base $a,$ the individual digits of the base-$a$ number will sometimes come out to multiple digits in base $b,$ but the idea is the same; just add them to the intermediate results.
Best Answer
In exactly the same way. Complements are involutions.
Look at the leading digit. 0..4 => positive; 5..9 => negative.