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

Answer: (D)

$(2z-1)(2\bar{z}-1)+(2\omega -1)(2\bar{\omega }-1)$
$\left[4|z{|}^2-2(z+\bar{z})+1\right]+\left[4|\omega {|}^2-2(\omega +\bar{\omega })+1\right]$
$E=6-4\mathrm{Re}(z)+4\omega \bar{\omega }-4\mathrm{Re}\omega$
$\text{ Now }|\omega -1{|}^2=1\Rightarrow (\omega -1)(\bar{\omega }-1)=1$
$\omega \bar{\omega }-(\omega +\bar{\omega })=0$
$\omega \bar{\omega }=\omega +\bar{\omega }$
$\omega \bar{\omega }=2\mathrm{Re}\omega$
$\text{ Hence }E=6-4\mathrm{Rez}+4\mathrm{Re}\omega$
$\text{ [Using }\omega \bar{\omega }=2\mathrm{Re}\omega \text{ ] }$
$=6+4(\mathrm{Re}\omega -\mathrm{Re}z)$
$=6+4\left(x_2-x_1\right)$
$E_{\max }=6+4(2+1)=18$
$E_{\min }=6+4(0-1)=2\Rightarrow \text{ say }[2,18]$