Minimum and Maximum value of |z| This is a question that I came across today:

If $|z-(2/z)|=1$...(1) find the maximum and minimum value of |z|, where z represents a complex number.

This is my attempt at a solution:

Using the triangle inequality, we can write:
$||z|-|2/z||≤|z+2/z|≤|z|+|2/z|$
Let $|z|=r$ which implies that $|r-2/r|≤1≤r+2/r$ (From (1))

How must I proceed to find the value of |z|? Please help! Much thanks in advance :)
 A: Since $|z-2/z| = 1$, we have
$$\begin{aligned}
1 = |z- 2/z|^2 &= (z - 2/z)(\overline{z} - 2/\overline{z}) \\
&= |z|^2 - 2z/\overline{z} - 2\overline{z}/z + 4/|z|^2
\end{aligned}$$
Write $z = re^{i\theta}$. Then $\overline{z} = re^{-i\theta}$. Substituting into the above, we obtain
$$\begin{aligned}
1 &= r^2 - 2e^{i2\theta} - 2^{-i2\theta} + 4/r^2 \\
&= r^2 - 4\cos(2\theta) + 4/r^2 \\
\end{aligned}$$
We can rearrange this to get
$$r^2 + 4/r^2 \leq 4\cos(2\theta) + 1 \leq 5$$
or equivalently,
$$r^4 -5r^2 + 4 \leq 0$$
Factor the left hand side to obtain
$$(r-1)(r+1)(r-2)(r+2) \leq 0$$
As $r$ must be positive, this means that $r+1$ and $r+2$ are positive. Therefore, $r-1$ and $r-2$ must have opposite signs (or one of them is zero), which forces $1 \leq r \leq 2$.
You can check that $z=1$ and $z=2$ are solutions to the original equation, so $1$ and $2$ are the minimum and maximum values of $|z|$.

Here is a picture showing the set of solutions. From this, we might be tempted to speculate that we are looking at two circles of radius $1/2$, centered at $z=3/2$ and $z=-3/2$. But this is not the case. To see this, consider the point $z = 3/2 + i(1/2)$. This point is on the circle of radius $1/2$ centered at $z=3/2$. However, it does not satisfy the given equation:
$$\begin{aligned}
|z-2/z| &= |3/2 + i(1/2) - 2/(3/2 + i(1/2))| \\
&= |3/2 + i(1/2) - 6/5 + i(2/5)| \\
&= |3/10 + i(9/10)| \\
&= \sqrt{9/10} \neq 1
\end{aligned}$$
I don't think it's an ellipse, either, since these are of the form $|z-a| + |z-b| = c$. If there is a way to transform $|z-2/z| = 1$ into that form, it's not obvious to me (also, our figure has not one but two "ellipses").

A: Using this equality we can find both minimum and maximum.
$$\left|z-\frac2z \right|\ge \left||z|-\frac2{|z|}\right|$$
$$\left||z|-\frac2{|z|}\right| \le 1$$
$$-1 \le |z|-\frac2{|z|}\le 1$$
$$|z|-\frac2{|z|}\ge -1$$
$$\left(|z| +\frac12\right)^2\ge \frac94$$
$$|z| \ge 1$$
$$|z|-\frac2{|z|}\le 1$$
$$\left(|z| - \frac12\right)^2\le \frac94$$
$$|z| \le 2$$
$$\implies 1\le |z|  \le 2$$
A: Using Complex Inequalities,   $$||z|-|w||\le|z+w|\le|z|+|w|$$
$w=-\dfrac2z$ and writing $|z|=r$
$$\left|r-\dfrac2r\right|\le\left|z-\dfrac2z\right|\le r+\dfrac2r$$
$$\implies\left|r-\dfrac2r\right|\le1\le r+\dfrac2r$$
Now as $r>0,$ $$\dfrac{r+\dfrac2r}2\ge\sqrt{r\cdot\dfrac2r}=\sqrt2\iff r+\dfrac2r\ge2\sqrt2>1$$
So, we need $$\left|r-\dfrac2r\right|\le1\iff-1\le r-\dfrac2r\le1$$
$$r-\dfrac2r\le1\iff r^2-r-2\le0$$
Now we know $(x-a)(x-b)\le0$ with $a\le b;a\le x\le b$
Here $-1\le r\le2$ But $r>0\implies0<r\le2$
Can check for $-1\le r-\dfrac2r$  to find $r\ge1$
So, the final range $$1\le r\le2$$
