Let $z = \cos \theta + i\sin \theta $ $z = \cos \theta + i\sin \theta $ $\mathop \sum \limits_{m = 1}^{15} {\rm{Im}}\left( {{z^{2m - 1}}} \right)\,{\rm{at}}\,\,\theta = {2^0}$ is

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

Let

z = \cos \theta + i\sin \theta

z = \cos \theta + i\sin \theta

\sum \limits_{m = 1}^{15} \mathrm{Im}\left( {{z^{2m - 1}}} \right)\,\mathrm{at}\,\,\theta = {2^0}

View answer and explanation

Correct option: D

\frac{1}{{4\sin {2^0}}}

Option D (

\frac{1}{{4\sin {2^0}}}

) is correct because the summation of the imaginary parts of the complex number series simplifies to this expression using trigonometric identities.

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