Greatest value of $|z|$ given that $\left|z- \frac 4z \right| = 2$. Why does using triangle inequality in this way not work?

Question

Greatest value of $|z|$ given that $\left|z- \dfrac 4z \right| = 2$ is?

This has been asked here before but my question is about a specific method that doesn't seem to work.

The method is:

$\left|z- \dfrac 4z \right|\le |z|+ \left|\dfrac{4}{z}\right|\implies 2 \le |z|+ \dfrac{4}{|z|}$

$\implies |z|^2 - 2|z| +4\ge 0$ since $|z| > 0$ (here)

which is true for all values of $|z|$.

Hence $|z| \in (0, \infty)$.

Then why is the maximum given by $|z| =1 + \sqrt 5$, in other words, what's the fault in this method ?

A user has also asked that in one of the comments but hasn't received a reply there.

Answer 1

You can apply the triangle inequality this way: $2 \geq |z-\frac{4}{z}| \geq |z|-|\frac{4}{z}|$, whence $|z|^2-2|z|-4 \leq 0$ whence $|z| \leq 1 + \sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+\sqrt{5}$

Answer 2

I was solving a similar problem in this question

edit

Assume $z\neq 0.\;$
$\left|z-\frac{4}{z}\right|$ is the distance of points representing $z$ and $\frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, \frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.

Solution

• If $z$ is real, then $z, \frac 4z$ lie on the same half-line starting in $0$ and we have $$\left|z-\frac 4z \right|=|z|-\frac {4}{|z|}\quad \text{or} \quad \left| z-\frac 4z \right|=\frac{4}{|z|}-|z|,$$ and so $$2=|z|-\frac {4}{|z|} \quad \text{or} \quad 2=\frac{4}{|z|}-|z|.$$

Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=\sqrt 5 +1\;$ from the first and $|z|=\sqrt 5 -1\;$ from the second one. Note that $\sqrt 5 -1=\frac{4}{\sqrt 5 +1}.$

• If $z=ib, b\in \mathbb{R},$ then $0$ lies on the segment with bounds $z, \frac 4z$. The equation to solve is then $2=|z|+\frac {4}{|z|}$ and doesn't have solution.

• In all other cases, by triangle inequality is $|z|<\sqrt 5 +1.$

The maximal value of $|z|$ is $\sqrt5 + 1,$ the only convenient numbers are $z=\pm(\sqrt 5 +1).$