Complex analysis, deceptively tricky problem.

[turnpages]

1. Let z be a complex variable. Describe the set of all z satisfying |z^2-z|<1.[\b]


I have a `brute force' solution, but it's really messy. Without a graphing utility, it would be nearly impossible to graph.

I just computed |z^2-z| in terms of x and y, and solved |z^2-z|=1 in this setting. The points inside this curve, then, satisfy the inequality.

It seems like I'm missing a more elegant solution, but I can't see it.

I thought I might be onto something when I did a change of variables and put the expression into the form |u^2-1/4|<1, but that didn't seem to help much.

 

[Dickfore]

Solve:
$$z^{2} - z = w$$
w.r.t. $z$. You get the multiple valued function:
$$z = \frac{1 + (1 + 4 w)^{\frac{1}{2}}}{2} = \left(w + \frac{1}{2}\right)^{\frac{1}{2}} + \frac{1}{2}$$
This can be expressed as sequence of two simple mappings:
$$u = w + \frac{1}{2}$$
$$v = u^{\frac{1}{2}}$$
$$z = v + \frac{1}{2}$$

1. What region in the w-plane doese $|w| < 1$ depict?
2. What is the mapping from the w to the u plane?
3. What is the mapping from the u plane to the v plane? Don't forget that $z^{1/2}$ is a double valued function.
4. What is the mapping from the v plane to the z plane?