≤ft({{6}-{2}}{4}+{{6}+{2}}{4}i)²020}=

Correct Answer is : 12−32i1/2-{{3}}{2}i21−23i

Correct Answer is : 12−32i1/2-{{3}}{2}i21−23i

Test Index

AP EAPCET 20-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 13 of 160

Marks: +1, -0

\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4}i\right)^{2020}=

Solution:

\left(\sqrt{2}\frac{\sqrt{3}-1}{2\cdot2}+i\sqrt{2}\frac{\sqrt{3}+1}{2\cdot2}\right)^{2030}

=\left(\frac{\frac{\sqrt{3}}{2}-\frac{1}{2}}{\sqrt{2}}+i\frac{\frac{\sqrt{3}}{2}+\frac{1}{2}}{\sqrt{2}}\right)^{2022}

=\left(\frac{\frac{1}{\sqrt{2}}\left(\frac{\sqrt{i}}{2}-\frac{1}{2}\right)}{1}+\frac{i\frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}\right)}{1}\right)^{2020}

=\left(\left(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\cdot\frac{1}{2}\right)+i\left(\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}+\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{2}\right)\right)\right)^{2000}

=(\cos 45^{\circ}\cos 30^{\circ}-\sin 45^{\circ}\sin 30^{\circ})

+i(\sin 45^{\circ}\cos 30^{\circ}+\cos 45^{\circ}\sin 30^{\circ}))^{2200}

=(\cos 75^{\circ}+i\sin 75^{\circ})^{2020}

=\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right)^{2020}

=\left(\cos\frac{5\pi}{12}\times2020+i\sin\frac{5\pi}{12}\times2020\right)

[De-moivre]

=\left(\cos\frac{2525\pi}{3}+i\sin\frac{2525\pi}{3}\right)

=(\cos(2525\times60)^{\circ}+i\sin(2525\times60)^{\circ}

=(\cos(n\pi-60)^{\circ}+i\sin(n\pi-60)^{\circ})

, where

n=842

=((-1)^{n}\cos60^{\circ}+i(-\sin60^{\circ}))

=(\cos60^{\circ}-i\sin60^{\circ})[\because n=

even

]

=\frac{1}{2}-i\frac{\sqrt{3}}{2}

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