Complex Numbers Ques 25
The inequality $|z-4|<|z-2|$ represents the region given by
(1982, 2M)
(a) $\operatorname{Re}(z) \geq 0$
(b) $\operatorname{Re}(z)<0$
(c) $\operatorname{Re}(z)>0$
(d) None of these
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Answer:
Correct Answer: 25.(d)
Solution:
Formula:
- Given, $|z-4|<|z-2|$
Since, $\left|z-z_1\right|>\left|z-z_2\right|$ represents the region on the right side of the perpendicular bisector of $z_1$ and $z_2$.
$ \begin{array}{ll} \therefore & |z-2|>|z-4| \\ \Rightarrow & \operatorname{Re}(z)>3 \text { and } \operatorname{Im}(z) \in \mathbb{R} \end{array} $
BETA
