Question

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

• A. $[1,9]$
• B. $[2,6]$
• C. $[2,12]$
• D. $[2,18]$

Answer 1

The correct option is D $[2,18]$

Least distance and greatest distance of any $z$ and $\omega$ from the point $(\frac{1}{2},0)$ are $\frac{1}{2}$ and $\frac{3}{2}$ respectively.


$\therefore {\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2\le {|z-\frac{1}{2}|}^2+{|\omega -\frac{1}{2}|}^2\le {\left(\frac{3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2$
Hence, $2\le |2z-1{|}^2+|2\omega -1{|}^2\le 18$