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  4. [(1+ cos (π / 8)+i sin (π / 8)/1+ cos (π / 8)-i sin (π / 8))]8 is equal to

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Q. $\left[\frac{1+\cos (\pi / 8)+i \sin (\pi / 8)}{1+\cos (\pi / 8)-i \sin (\pi / 8)}\right]^8$ is equal to

Complex Numbers and Quadratic Equations

Solution:

$\left[\frac{1+\cos (\pi / 8)+i \sin (\pi / 8)}{1+\cos (\pi / 8)-i \sin (\pi / 8)}\right]^8=\left(\frac{2 \cos ^2 \frac{\pi}{16}+i \cdot 2 \sin \frac{\pi}{16} \cos \frac{\pi}{16}}{2 \cos ^2 \frac{\pi}{16}-i 2 \cos \frac{\pi}{16} \sin \frac{\pi}{16}}\right)^8$
$=\left(\frac{\cos \frac{\pi}{16}+i \cdot \sin \frac{\pi}{16}}{\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}}\right)^8=\cos \pi+i \sin \pi=-1$