Question

If $\left|z\right|=1$ and $\left|\omega -1\right|=1$ where z, $\omega$, $\in$ C, then the value of set of ${\left|2z-1\right|}^2+{\left|2\omega -1\right|}^2$:

• A. $(1,9)$
• B. $(2,6)$
• C. $(2,12)$
• D. $(2,18)$

Answer 1

The correct option is D $(2,18)$

$|z|=1$ implies that $z$ is a circle with radius $1$ unit with a center

$(0,0)$ in the complex plane and

$|\omega -1|=1$ implies that it is a circle with the radius $1$ unit center

$(1,0)$ in the complex plane.

We can write

$|2z-1{|}^2+|2\omega -1{|}^2$

as

$4(|z-\frac{1}{2}{|}^2+|\omega -\frac{1}{2}{|}^2)$

It means it is the range of $4$ times the sum of the squares of the distances of points from $(0.5,0)$ from the respective circles of locus of the $z,\omega$.

The point $(0.5,0)$ is 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 anyone circle's range.

The min or max possible values of the distance from the point becomes the distances between the point and the vertices of the diameter on which it lies that is

$\frac{1}{2},\frac{3}{2}$

$8\times ((\frac{1}{2}{)}^2,(\frac{3}{2}{)}^2)$

$(2,18)$