JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

\alpha=\cos \left(\frac{8 \pi}{11}\right)+i \sin \left(\frac{8 \pi}{11}\right)

, then

\operatorname{Re}\left(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5\right)

is equal to

View answer and explanation

Correct option: B

-\frac{1}{2}

Option B (-

\frac{1}{2}

) is correct because it is the real part of the sum of complex numbers on the unit circle.

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\left|\sum\limits_{r=1}^n z_r\right|

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\alpha, \beta

\gamma

|\alpha-\beta|

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z \in C

2x = - 1 + \sqrt 3 i

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