If $|z|=1$ and $|\omega -1| =1$ where $z, \omega \in C$, then the largest set of values of $|2z - 1|^2 + | 2\omega -1|^2$ equals:

$|z|=1$ implies that Z in Q circle with radius with a center (0,0) in the complex plane.

$|w-1|=1$ implies that it in a circle with the radius unit, center $(1,0)$ in the complex plane.

$|2z-1|^2+|2w-1|^2=4\left(\left|z-\dfrac{1}{2}\right|^2+\left|w-\dfrac{1}{2}\right|^2\right)$

It means it is the range of $4$ times the sum of the square of the distance of points from $(0.5,0)$ from the respective circles of low of the z,w

The point $(0.5,0)$ in symmetric with the both circles on the center of a radius on the X-axis so both of them get equal ranges so it simply becomes $8$ times of annual circle range.

The min or max possible values of the distance from the point become the distance b/w the point and the vertices of the diameter on choice it lies that is

$\rightarrow \dfrac{1}{2},\dfrac{3}{2}$

$=8\times \left[\left(\dfrac{1}{2}\right)^2,\left(\dfrac{3}{2}\right)^2\right]=[2,18]$