(1+i 2 )^8+ (1-i 2 )^8= - Vidyadip

MCQ

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

A
1
✓
2
C
4
D
8

✓

Answer

Correct option: B.

2

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

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