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.
Remember the meaning of base 10 notation; when you write a number as
$$d_nd_{n-1}\cdots d_2d_1d_0$$
where $d_i$ is the $i$th digit (from right to left), what you are saying is that the number is equal to:
$$ d_0\times 10^0 + d_1\times 10^1 + d_2\times 10^2 + \cdots + d_n\times 10^n.$$
So, for example, $5381$ represents the number
$$1\times 10^0 + 8\times 10^1 + 3\times 10^2 + 5\times 10^3 = 1 + 80 + 300 + 5000.$$
Writing a number in binary (base 2) is meant to represent the number in exactly the same way, but with powers of $2$ in stead of powers of 10: the expression
$$e_k\cdots e_3e_2e_1e_0{}_{[2]}$$
represents the number
$$n=e_0\times 2^0 + e_1\times 2^1 + e_2\times 2^2 + e_3\times 2^3 + \cdots + e_k\times 2^k.$$
Since every summand except the first one is a multiple of $2$, we can write:
$$\begin{align*}
n&=e_0\times 2^0 + e_1\times 2^1 + \cdots + e_k\times 2^k\\
&=e_0 + \left(e_1\times 2\right) + \left(e_2\times 4\right) + \cdots \left(e_k\times 2^{k}\right)\\
&= e_0 + \left( 2 \times e_1\right) + \left( 2\times (e_2\times 2)\right) + \cdots + \left(2\times(e_k\times 2^{k-1})\right)\\
&= e_0 + 2\Bigl(e_1 + (e_2\times 2) + \cdots + (e_k\times 2^{k-1})\Bigr).
\end{align*}$$
That means that when you divide $n$ by $2$ you get a remainder of $e_0$ (the right-most digit of the base 2 expression of $n$), and a quotient of
$$q_1=e_1\times 2^0 + e_2\times 2^1 + \cdots + e_k\times 2^{k-1}.$$
Now you can determine the next binary digit of $n$ by repeating the process with $q_1$: we write
$$q_1 = e_1 + 2\Bigl(e_2 + e_3\times 2 + \cdots + e_k\times 2^{k-2}\Bigr),$$
so the remainder of dividing $q_1$ by $2$ is the penultimate digit of the binary expression of $n$, and the quotient is $q_3$, with
$$q_3 = e_2 + e_3\times 2 + \cdots + e_k\times 2^{k-2}.$$
Lather, rinse, and repeat until the remainder quotient is $0$.
Best Answer
HINT:
Let's consider the case of the Gaussian integers of the form $n+im$ and let the base be $-1+i$.
In general, if one has a base and a number to expand in that base then the one will divide the number repeatedly by the base and will take notes of the remainders. I claim that in the case of the base $-1+i$ the remainder is $0$ or $1$ if the number to be expanded is a Gaussian integer.
So, take a Gaussian integer $n+im$ and divide it by the base chosen:
$$\frac{n+im}{-1+i}=\frac{(n+im)(-1-i)}{2}=\frac{m-n}{2}-i\frac{n+m}{2},$$ where the nominator and the denominator have been multiplied by $-1-i$ (the complex conjugate of the base).
If both $n$ and $m$ are even or both of them are odd then $m-n$ and $m+n$ both are even and the result of the division is a Gaussian integer, $$\frac{m-n}{2}-i\frac{n+m}{2}$$
and the remainder is $0$.
If only one of $n$ or $m$ is even (the other one is odd) then one can consider the following version of the quotient:
$$\frac{n+im}{-1+i}=\frac{m-n+1}{2}-i\frac{n+m-1}{2}-\frac{1}{2}(1+i).$$ In this case the result of the division is the following Gaussian integer:
$$\frac{m-n+1}{2}-i\frac{n+m-1}{2}$$ and the remainder is $1$. Indeed $$\big(-1+i\big)\big( \frac{m-n+1}{2}-i\frac{n+m-1}{2} \big)=n+im-1.$$
One can repeat this prescription with the resulting Gaussian integers until, as a result of the repeated divisions by $2$, the original number disappears and the sequence of the remainders stays...
Let's illustrate the algorithm by the example given in the OP: $$24-i11.$$
If $$(m-n)-i(n+m)=(24-(-11))-i(24+(-11)=-35-i13$$
(See the second pair of columns in the table below) was divisible by $2$ then the third pair of columns would contain the same without any modification and the remainder would be zero (See the column called Remainder). Since $-35-i13$ is not divisible by two, there is a modification according to the formula $$(m-n+1)-i(n+m-1)=-34-i12$$ and the remainder is $1$. The first red arrow points at the result of the first division. (Division by $2$ is performed now.) Then the procedural step detailed above will be repeated until the corresponding value in the third pair of columns has become $0+i0$.
$ \color{white}{bbi}$
If one multiplies the remainders by the different powers of the base $i-1$ and sums up the results then the result will be the complex number which was to be expanded:
$ \color{white}{bbbbbbbbb}$
Finally, we have $$(24-i11)_{10} =(110010110011)_{-1+i}.$$
I hope that this hint above will help the OP solve the other problems. This is not impossible since the skeleton of the method is the same as that of the one that we use in case of the simplest base conversions in "the every day life."