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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.

SihOASHoihd
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  • I've removed the "logic" tag, which is for questions about the specific subfield mathematical logic, not questions about mathematical reasoning in general. – Noah Schweber Jun 19 '17 at 18:24
  • I agree this is more about general reasoning. A possible reason for keeping the "logic" tag: any systemic approach to these types of questions would surely involve formal logic. – SihOASHoihd Jun 19 '17 at 19:01

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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.

dxiv
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    I waited before looking at your answer to see if I could figure it. I couldn't. Thanks for the edit. – SihOASHoihd Jun 20 '17 at 23:22
  • @SihOASHoihd Here are some more problems among the recently posted, where you could practice similar ideas to solve without resorting to Cartesian - 1, 2, 3, 4, 5, 6. – dxiv Jun 21 '17 at 00:30
  • Thanks, I solved those problems. I think my issue was that I was stubbornly trying to only use algebraic manipulations. The way I see it, your key step of noticing that you could just look at the numerator was not a purely algebraic step: it required some reasoning that was extra-algebraic. In this sense, I think I was originally correct in saying that such a manipulation is impossible. (This is why I kept bringing up logic; how does one define a purely algebraic move?) Nevertheless I didn't try your approach and I am happy you showed it to me. – SihOASHoihd Jun 23 '17 at 17:53