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 $2\le |2z-1{|}^2+|2\omega -1{|}^2\le 18$