Question

Let $z$ be any point in complex plane and let $w$ be any point satisfying $|w-2-i|<3.$ Then, $\left|z\right|-\left|w\right|+3$ lies between

• A. $-6$ and $3$
• B. $-3$ and $6$
• C. $-6$ and $9$
• D. $-3$ and $9$

Answer 1

Answer: (D)

$-3$ and $9$

$|w-2-i|<3$ $|\left|z\right|-\left|w\right||<|z-w|$
$|z-w|=$ distance between $z$ and $w,z$ is fixed.

It will be maximum for diametrically opposite points.
$\therefore |z-w|<6\therefore -6<\left|z\right|-\left|w\right|<6$

Add 3 in all.
$\therefore -3<\left|z\right|-\left|w\right|+3<9$