If |z-2/z| =2 then prove that √3-1<=|z|<=√3+1?

z - 2/z = 2
z^2 - 2 = 2z
z^2 - 2z = 2
z^2 - 2z + 1 = 2 + 1
(z - 1)^2 = 3
z - 1 = +/- sqrt(3)
z = 1 +/- sqrt(3)

z - 2/z = -2
z^2 - 2 = -2z
z^2 + 2z = 2
z^2 + 2z + 1 = 3
(z + 1)^2 = 3
z + 1 = +/- sqrt(3)
z = -1 +/- sqrt(3)

z = -1 - sqrt(3) , 1 - sqrt(3) , sqrt(3) - 1 , 1 + sqrt(3)

|z| = |-1 - sqrt(3)| , |1 - sqrt(3)| , |sqrt(3) - 1| , |1 + sqrt(3)|
|z| = 1 + sqrt(3) , sqrt(3) - 1 , sqrt(3) - 1 , 1 + sqrt(3)
|z| = sqrt(3) - 1 , sqrt(3) + 1

|z−2/z|=2 ⟹ |z²−2|=2|z| (z is complex)

By triangle inequality |z²−2|+|2| ≥ |z|² so |z|²−2|z|−2 ≤ 0
The roots of quadratic are 1±√3 hence 1−√3 ≤ |z| ≤ 1+√3 ⟹ |z| ≤ 1+√3 … (i)

Also by triangle inequality |z²−2|+|−z²| ≥ |−2| so |z|²+2|z|−2 ≥ 0
The roots of quadratic are −1±√3 hence |z| ≥ −1+√3 or |z| ≤ −1−√3 ⟹ |z| ≥ −1+√3 … (ii)

Combine (i) & (ii) for −1+√3 ≤ |z| ≤ 1+√3

I'm going to assume that you mean exactly what you wrote, and not | (z-2) / z | = 2.

Starting with |z-2/z| = 2, then either (1) z-2/z = +2, or (2) z-2/z = -2.
Just solve each one.
(1) z-2/z = 2, so then
z^2 - 2 = 2z ==> z = 1 - sqrt(3), or z = 1 + sqrt(3)

(2) z-2/z = -2, so then
z^2 - 2 = -2z ==> z = -1 - sqrt(3), or z = -1 + sqrt(3)

We have FOUR solutions: z = +/-1 +/- sqrt(3)