Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Prove the following

Question:

If $z=\frac{\sqrt{3}}{2}+\frac{i}{2}(i=\sqrt{-1})$

then $\left(1+i z+z^{5}+i z^{8}\right)^{9}$ is equal to

  1. -1

  2. 1

  3. 0

  4. $(-1+2 \mathrm{i})^{9}$


Correct Option: 1

Solution:

$z=\frac{\sqrt{3}}{2}+\frac{i}{2}=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}$

$\Rightarrow \mathrm{z}^{5}=\cos \frac{5 \pi}{6}+\mathrm{i} \sin \frac{5 \pi}{6}=\frac{-\sqrt{3}+\mathrm{i}}{2}$

and $z^{8}=\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}=-\left(\frac{1+i \sqrt{3}}{2}\right)$

$\Rightarrow\left(1+i z+z^{5}+\mathrm{iz}^{8}\right)^{9}=\left(1+\frac{\mathrm{i} \sqrt{3}}{2}-\frac{1}{2}-\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}-\frac{\mathrm{i}}{2}+\frac{\sqrt{3}}{2}\right)^{9}$

$=\left(\frac{1+\mathrm{i} \sqrt{3}}{2}\right)^{9}=\cos 3 \pi+\mathrm{i} \sin 3 \pi=-1$

Leave a comment

None
Free Study Material