If $\left|z+\frac{2}{z}\right |=2$, then prove that max value of $|z|$ is $\sqrt 3 + 1$ (Corrected version)

Question

Then the expression will be $$|z|+\frac{2}{|z|}\ge 2$$

Solving the quadratic inequality

$$|z|\ge1+i$$ and $$|z|\le1-i$$

I don’t know how to solve further. Please help

Thanks!

this is the corrected version of an identical question asked by me. I am sorry for the inconvenience caused

Answer 1

You’ve made a couple fundamental errors.

Firstly, $\geq$ and $\leq$ generally lose their meaning in complex numbers. Is $i>3$? $i-3$ greater than or less than $0$?

Secondly, $|z| = \text{a complex number}$ does not make sense either. The magnitude of a number is explicitly real. How long would it take for you to walk $5-12i\text{ meters}$? (And don’t say ${5-12i\over v}$ seconds. What were you doing $i$ seconds ago?)

So, the fact that (for notational brevity, $|z| = x$) $x+\frac 2x \geq 2\implies (x-1)^2\geq -1$ conveys no groundbreaking information at all— being the square of a real number, $(x-1)^2$ is positive so it is indeed $\geq-1$. In fact it is strictly $> -1$ since equality cannot be achieved.

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Instead note that

$$\left|z - \left(-\frac 2z\right)\right|\geq \left||z|-\left|\frac 2z\right|\right|$$

So if $|z|<\sqrt 2$ it is obviously $<\sqrt 3 +1$, and if $|z|\geq \sqrt 2$ we have:

$$|z|-\frac 2{|z|} \leq 2$$

$$\implies x^2-2x \leq 2$$ $$\implies (x-1)^2\leq 3$$ $$\implies \boxed{|z|\leq \sqrt 3 + 1}$$

Answer 2

$|z^{2}+2|=2|z|$ gives $|z|^{2}-2 \leq 2|z|$ or $(|z|-1)^{2} \leq 3$. Hence $|z| \leq 1+\sqrt 3$. I will let you find a value of $z$ for which equality holds.

You have to choose $z$ to be $(1+\sqrt 3)e^{i\theta}$ for some $\theta$.

Answer 3

Answer

Since $\left|z+\frac2z\right|=2$, $$\newcommand{\Re}{\operatorname{Re}} \begin{align} 4 &=\left(z+\frac2z\right)\left(\bar z+\frac2{\bar z}\right)\\[6pt] &=|z|^2+\frac4{|z|^2}+4\Re\left(\frac{\bar z}z\right)\tag1 \end{align} $$ which is equivalent to $$ \begin{align} \left(|z|-\frac2{|z|}\right)^2 &=-4\Re\left(\frac{\bar z}z\right)\\[6pt] &=-4\cos(2\arg(z))\tag2 \end{align} $$ That is, $$ -2\le|z|-\frac2{|z|}\le2\tag3 $$ Since $x-\frac2x$ is monotonically increasing for $x\gt0$, we get that $$ \bbox[5px,border:2px solid #C0A000]{\sqrt3-1\le|z|\le\sqrt3+1}\tag4 $$

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Visualization

We get equality in $(3)$ and $(4)$ when $\arg(z)=\pm\frac\pi2$, as is seen in $(2)$ and the plot of $$ |z|-\frac2{|z|}=\pm2\sqrt{-\cos(2\arg(z))}\iff r=\sqrt{2-\cos(2\theta)}\pm\sqrt{-\cos(2\theta)}\tag5 $$

Answer 4

Let $z=re^{i\theta}$. Then, $\left|z+\frac{2}{z}\right |=\left|r+\frac{2}{r}e^{-i2\theta}\right |=2$ leads to

$$\left(r+\frac2r \cos 2\theta \right)^2+\frac{4}{r^2}\sin^2 2\theta= 4$$

or,

$$r^2+\frac{4}{r^2}=8\sin^2\theta\le 8$$

Rearrange the inequality to get,

$$(r^2 -4 )^2\le 12$$

which leads to

$$r\le \sqrt{4+2\sqrt3}=1+\sqrt3$$