Solve this following

Question:

Let $\mathrm{z}=\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{\mathrm{i}}{2}\right)^{5}$. If $\mathrm{R}(\mathrm{z})$ and $\mathrm{I}[\mathrm{z}]$

respectively denote the real and imaginary parts of $z$, then :

 

 

 

  1. $\mathrm{R}(\mathrm{z})>0$ and $\mathrm{I}(\mathrm{z})>0$

  2. $\mathrm{R}(\mathrm{z})<0$ and $\mathrm{I}(\mathrm{z})>0$

  3. $R(z)=-3$

  4. $I(z)=0$


Correct Option: , 4

Solution:

$z=\left(\frac{\sqrt{3}+i}{2}\right)^{5}+\left(\frac{\sqrt{3}-i}{2}\right)^{5}$

$\mathrm{z}=\left(\mathrm{e}^{\mathrm{i} \pi / 6}\right)^{5}+\left(\mathrm{e}^{-\mathrm{i} \pi / 6}\right)^{5}$

$=\mathrm{e}^{\mathrm{i} 5 \pi / 6}+\mathrm{e}^{-\mathrm{i} 5 \pi / 6}$

$=\cos \frac{5 \pi}{6}+i \frac{\sin 5 \pi}{6}+\cos \left(\frac{-5 \pi}{6}\right)+i \sin \left(\frac{-5 \pi}{6}\right)$

$=2 \cos \frac{5 \pi}{6}<0$

$\mathrm{I}(\mathrm{z})=0 \quad$ and $\operatorname{Re}(\mathrm{z})<0$

Option (4)

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