2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum. $\endgroup$ – fleablood Dec 12 '18 at 20:50
  • $\begingroup$ "which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds. $\endgroup$ – Arthur Dec 12 '18 at 20:51
  • 1
    $\begingroup$ Basically that shows $|z| \le 1+\sqrt{5} \le \infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit. $\endgroup$ – fleablood Dec 12 '18 at 20:52
  • 1
    $\begingroup$ Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing. $\endgroup$ – fleablood Dec 12 '18 at 20:59
4
$\begingroup$

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 $z=1+\sqrt{5}$

$\endgroup$
1
$\begingroup$

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 $z=\pm(\sqrt 5 +1).$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.