Question

If $z$ satisfies $|z-1|<|z+3|$ then $\omega =2z+3-i$ (where $i=\sqrt{-1}$) satisfies

• A. $|\omega -5-i|<|\omega +3+i|$
• B. $|\omega -5|<|\omega +3|$
• C. $\mathrm{lm}\left(i\omega \right)>1$
• D. $|arg(\omega -1)|<\frac{\pi }{2}$

Answer 1

The correct option is D $|arg(\omega -1)|<\frac{\pi }{2}$

$|z-1|<|z+3|$

$\Rightarrow |z-\left(1\right)|<|z-(-3)|$

$\Rightarrow$ distance of a $pt-z$ from $(1+0i)$ is greater than its distance from $(-3+0i)$

$\Rightarrow z$ always remains on the left side of $-1$

$w=2z+3-i$

let $z=-3/2\Rightarrow w=-i$

$A)|\omega -5-i|<|\omega +3+i|$

$\Rightarrow |-5-2i|<|3|$

$B)|\omega -5|<|\omega +3|$

$\Rightarrow |-i-5|<|-i+3|$

$C\left)\mathrm{lm}\right(i\omega )>1$

$\Rightarrow lm(-i^2)=1>1$

$D\left)|arg\right(\omega -1)|<\frac{\pi }{2}$

$\Rightarrow |arg(-1-i)|<\pi /2$