Complex Numbers 2 Question 20
20. 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: 20. (b)
Solution:
- Given, $|z-4|<|z-2|$
Since, $\left|z-z _1\right|>\left|z-z _2\right|$ represents the region on right side of perpendicular bisector of $z _1$ and $z _2$.
$$ \begin{array}{ll} \therefore & |z-2|>|z-4| \\ \Rightarrow & \operatorname{Re}(z)>3 \text { and } \operatorname{Im}(z) \in R \end{array} $$
$$ \because \omega=\frac{-1+i \sqrt{3}}{2} \text { and } \omega^{2}=\frac{-1-i \sqrt{3}}{2} $$
Now, $\quad \frac{\sqrt{3}+i}{2}=-i \frac{-1+i \sqrt{3}}{2}=-i \omega$
$$ \text { and } \quad \frac{\sqrt{3}-i}{2}=i \frac{-1-i \sqrt{3}}{2}=i \omega^{2} $$
$\therefore \quad z=(-i \omega)^{5}+\left(i \omega^{2}\right)^{5}=-i \omega^{2}+i \omega$
$=i\left(\omega-\omega^{2}\right)=i(i \sqrt{3})=-\sqrt{3}$
$\Rightarrow \quad \operatorname{Re}(z)<0$ and $\operatorname{lm}(z)=0$
Alternate Solution
We know that, $\quad z+\bar{z}=2 \operatorname{Re}(z)$
If
$$ z=\frac{\sqrt{3}}{2}+\frac{i}{2}^{5}+\frac{\sqrt{3}}{2}-\frac{i}{2}^{5} \text {, then } $$
$z$ is purely real. i.e. $\operatorname{Im}(z)=0$
