<s>[INST] Maximum value of $|z|$ given $\lvert z-\frac 4z \rvert = 8$? The question is $$ \left|z-\frac 4z \right| = 8$$ Find the max value of $ |z|$
You know how the triangle inequality is:
$$ \bigg| | z_1 | - | z_2| \bigg| \leqq | z_1 \pm z_2 | \leqq | z_1 | + | z_2 | $$
The solutions used only the left hand side inequality, and also ignoring the absolute values outside of $ | z_1 | - | z_2 | $, i.e. they solved
$ | z |  - \left| \frac 4z \right| \leqq 8$ to obtain the answer $ | z |_{max} = 4 + 2 \sqrt{5} $
I am confused about this in two ways, firstly, the way they solved it aren't they assuming here that $| z | \geqq | \frac 4z |$ ? Also can you just ignore the right hand side inequality?
 [/INST] Redo:
We have three potential inequalities
$|z| - |\frac z4| \le ||z| - |\frac z4|| \le |z -\frac z 4|=8$ or

*

*$|z| -|\frac z4| \le 8$.

$-|z| + |frac z4| \le ||z| - |\frac z4|| \le |z -\frac z 4|=8$ or


*$|\frac z4| -|z| \le 8$.

$8=|z-\frac z4|\le |z| + |\frac z4|$ or


*$|z| +|\frac z4| \ge 8$.

All three of these equations are true for any possible value of $z$ where $|z +\frac 4z| =8$.
The first yields $|z|\le 4 + 2\sqrt 5$ AND $|z|\ge 2\sqrt 5-4$ so $2\sqrt 5-4\le |z|\le 4+2\sqrt 5$.  That is always true.
The second yields $|z|\ge 2\sqrt 5-4$ OR that $|z|\le -4 -\sqrt 5$ but that is impossible.  So $|z|\ge 2\sqrt 5 -4$.  That is always true.
The third yields $|z| \ge  2\sqrt 3$ OR $|\le 4-2\sqrt 3$.
Putting these results together and noting that $2\sqrt 5-4 \le 4-2\sqrt 3 \iff \sqrt 5 + \sqrt 3 < 4 \iff 5+2\sqrt {15}+3 < 16\iff 8+2\sqrt {15}< 16=8+2\sqrt {16}$ which it is we get.
$2\sqrt 5 - 4\le |z| \le 4-2\sqrt 3$ or $4+2\sqrt 3 \le |z| \le 4+ 2\sqrt 5$
So $|z|_{max} = 4+2\sqrt 5$.
As we were only asked for the maximum value and we get that from inequality 1), inequality 1) is the only one we need to consider.
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