Complex Numbers Ques 1

  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. $\operatorname{Im}(z)=0$



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