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

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 ?

• 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. – fleablood Dec 12 '18 at 20:50
• "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. – Arthur Dec 12 '18 at 20:51
• 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. – fleablood Dec 12 '18 at 20:52
• 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. – fleablood Dec 12 '18 at 20:59

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 $$z=1+\sqrt{5}$$

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