TITLE: Least and greatest value of $|z|$ given that $|z- 4/z| = 2$?
QUESTION [5 upvotes]: If the complex number $z$ satisfies the equation $\left\lvert z- {4\over z}\right\rvert = 2$ then the least and the greatest values of $|z|$ are ?
My try
 $\left\lvert z- {4\over z}\right\rvert = 2$ 
$\left| |z| - \left\lvert z- {4\over z}\right\rvert \right| \le 2$.

REPLY [8 votes]: Given $$\left|z-\frac{4}{z}\right| = 2$$ and here we have to find $\max$ and $\min$ of $|z|$
So $$|z| = \left|\left(z-\frac{4}{z}\right)+\frac{4}{z}\right|\leq \left|z-\frac{4}{z}\right|+\frac{4}{|z|}$$
Above we have used $\bf{\triangle \; Inequality}$
So $$|z|\leq 2+\frac{4}{|z|}\Rightarrow |z|^2-2|z|\leq 4$$
So $$\left(|z|-1\right)^2 \leq 5$$
Now after that You can solve it.

REPLY [6 votes]: If you know $\left|\frac zw\right|=\frac{|z|}{|w|}$, then you can solve your inequality for real $|z|$.  Then see if real $z$ attains those max and min.
$$|r-4/r|\leq2\\-2r\leq r^2-4\leq2r\\5\leq(r+1)^2,(r-1)^2\leq5\\-1+\sqrt{5}\leq r\leq1+\sqrt{5}$$