Complex Numbers Ques 1
- If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5$, then
(1982, 2M)
(a) $\operatorname{Re}(z)=0$
(b) $\operatorname{Im}(z)=0$
(c) $\operatorname{Re}(z)>0, \operatorname{Im}(z)>0$
(d) $\operatorname{Re}(z)>0, \operatorname{Im}(z)<0$
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Answer:
Correct Answer: 1.(b)
Solution: (b) Given, $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5$
$ \left[\because \omega=\frac{-1+i \sqrt{3}}{2} \text { and } \omega^2=\frac{-1-i \sqrt{3}}{2}\right] $
Now, $\quad \frac{\sqrt{3}+i}{2}=-i\left(\frac{-1+i \sqrt{3}}{2}\right)=-i \omega$
and $\quad \frac{\sqrt{3}-i}{2}=i\left(\frac{-1-i \sqrt{3}}{2}\right)=i \omega^2$
$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} $
$\Rightarrow \operatorname{Re}(z) <0 \text { and } \operatorname{lm}(z)=0$
Alternate Solution
We know that, $\quad z+\bar{z}=2 \operatorname{Re}(z)$
If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5$,
then $z$ is purely real. i.e. $\operatorname{Im}(z)=0$
BETA
