Update

I haven't posted anything here for nearly two years - where does the time go! Since that time I have continued on with my study of number theory and getting through chapters 9 and 10 of John Stillwell's Elements of Number Theory on Quadratic Reciprocity and Rings. I was making very slow progress I must say. I started chapter 11 on Ideals but before I had got very far I felt like a change and last September began reading the books in the OU's M337 Complex Analysis course. So far I have only worked through book A1 and am halfway through A2. Slow progress indeed! I just do enough to keep the old brain ticking over.

Book A1 of M337 introduces complex numbers and goes over some familiar stuff that we have seen in MS221 and M208 (e.g. nth roots of a complex number). Book A2 is a bit more interesting as it starts to deal with complex functions where the domain and image are sets of complex numbers. Here, nothing is very obvious as we are mapping from a 2d plane to a 2d plane and it is not easy to see what a function is doing. One thing that struck me, that I hadn't really thought about before, is that the complex plane is not an ordered field like the set of real numbers. For example, you can't write an inequality like 1+2i > 1+i. So any ordering can only be done with functions that map to a real line (such as the modulus).

I also liked the approach to determining the image of a complex function. For example, Problem 1.2(b) asks you to determine the image of the function f with rule f(z) = (3z+1)/(z+i). Now there is a convention which states that if a function is specified by just its rule then it is understood that its domain is the set of all complex numbers to which the rule is applicable and its codomain is C (the set of all complex numbers). So here the domain A = C - {-i} (i.e. the whole of the complex plane except the point -i) because when z = -i the denominator in the rule is zero. So the image of f is

$$ f(A) =  \{ w = \frac{3z+1}{z+i} : z \in A \}$$

The aim is to write this set in terms of the image point w and a condition on w. Since

$$  w = \frac{3z+1}{z+i} \Leftrightarrow w(z+i)=3z+1 \Leftrightarrow  z(w-3)=1-wi \Leftrightarrow z = \frac{1-wi}{w-3}$$

the the image of f is

$$ f(A) =  \{ w : z = \frac{1-wi}{w-3} \neq -i \}$$

Now we can imagine the image point w moving around the complex plane and at the same time see the point z in the domain from which it is mapped. We want both w to exist and z not to be equal to -i. You can see that z is never equal to -i for any w in C. Also, (1-wi)/(w-3) is just another complex number and so, as w ranges over all of C, so does z. The only point that is excluded in the image is when w=3 as again this would make the denominator in z zero. Thus we have that

$$ f(A) =  \{ w : w \neq 3 \} $$

and so the image of f is C - {3}. Neat!