Q:

Soft question about manipulating complex inequalities

I am sometimes surprised by how adept people are at manipulating inequalities involving absolute values of complex numbers.
I know that by representing z in rectangular coordinates, it is relatively straightforward to show that  $\mid \frac{1}{z} -1 \mid = \frac{1}{2}$ is the circle $\mid z- \frac{4}{3} \mid = \frac{2}{3}$ (using basic properties of the transformation $w= \frac{1}{z}$ also works I believe).  What I don't have a good intuition for is whether or not the former expression can be algebraically manipulated into the latter expression by staying in the variable z (i.e., not switching to rectangular or polar coordinates).  After playing around with this for a little, my guess is that this is not possible.  What I am wondering is: are there any general rules that I could apply at the beginning to tell me whether or not such a manipulation is possible.    

A:

The condition can be rewritten as:
$$
0 = \left| \frac{1}{z} -1 \right|^2 - \frac{1}{4}= \frac{(1-z)(1-\bar z)}{|z|^2} - \frac{1}{4} = \frac{4 - 4 z - 4 \bar z+4|z|^2 - |z|^2}{4 |z|^2}
$$
It follows that the numerator must be $0\,$, so:
$$
0 = 4 - 4 z - 4 \bar z+3|z|^2 = \frac{1}{3}\Big((3z-4)(3 \bar z -4) - 4\Big) = \frac{1}{3}\Big(|3z-4|^2 - 4\Big)
$$
The latter gives $\displaystyle|3z-4|=2 \iff \left|z - \frac{4}{3}\right| = \frac{2}{3}\,$.

[ EDIT ]  The technique is applicable in many cases of such transformations. The key points are:

$\;|z|^2 = z \bar z\,$, which allows expanding the modulus in terms of conjugates;
 recognizing that $|z|^2 + \bar a z + a \bar z = (z+a)(\bar z + \bar a) - |a|^2 = |z+a|^2 - |a|^2\,$, which allows collecting such expressions back into a modulus form.

