Question

The value of ${\left(\frac{1+i}{\sqrt{2}}\right)}^8+{\left(\frac{1-i}{\sqrt{2}}\right)}^8$ is equal to

• A. $4$
• B. $6$
• C. $8$
• D. $2$

Answer 1

The correct option is D $2$

We have ${\left(\frac{1+i}{\sqrt{2}}\right)}^8+{\left(\frac{1-i}{\sqrt{2}}\right)}^8$

$={[\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}]}^8+{[\cos \frac{\pi }{4}-i\sin \frac{\pi }{4}]}^8$

$=\cos 2\pi +i\sin 2\pi +\cos 2\pi -i\sin 2\pi$ [by de-moivre's theorem]

$=2\cos 2\pi =2\left(1\right)=2$