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  4. The value of ((1+i/√2))8+((1-i/√2))8 is equal to

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

Complex Numbers and Quadratic Equations

Solution:

We have, $\left(\frac{1+i}{\sqrt{2}}\right)^{8}+\left(\frac{1-i}{\sqrt{2}}\right)^{8}$
$=\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right]^{8}+\left[\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right]^{8}$
$=\cos 2 \pi+i \sin 2 \pi+\cos 2 \pi-i \sin 2 \pi$
$=2 \cos 2 \pi=2(1)=2 $
$[$ By De-Moivre's theorem $]$