Question

If $z$ satisfies $|z+1|<|z-2|,$ and $\omega =3z+2+i,$ then

• A. $|\omega +1|<|\omega -8|$
• B. $|\omega +1|<|\omega -7|$
• C. $\omega +\overline{\omega }>7$
• D. $|\omega +5|<|\omega -4|$

Answer 1

The correct option is A $|\omega +1|<|\omega -8|$

$|z+1|<|z-2|$ [ Given ]

$w=3z+2+i$

$w+\overline{w}=3(z+\overline{z})+4$

$|z+1{|}^2<|z-2{|}^2$

$z\overline{z}+z+\overline{z}+1<z\overline{z}-2z-2\overline{z}+4$

$\Rightarrow$ $z+\overline{z}<1$

$\Rightarrow$ $w+\overline{w}<7$

Now taking option a,

$|w+1{|}^2<|w-8{|}^2$

$\Rightarrow$ $w\overline{w}+w+\overline{w}+1<w\overline{w}-8w-8\overline{w}+64$

$\Rightarrow$ $9(w+\overline{w})<63$

$\Rightarrow$ $(w+\overline{w})<7$

$\therefore$ We can see option $A$ satisfies.