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JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy
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Application +4 | -1
If z+1z=2cos⁡θ,z∈Cz+\frac{1}{z}=2 \cos \theta, z \in \mathbf{C} then z2n−2zncos⁡(nθ)z^{2 n}-2 z^n \cos (n \theta) is equal to
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Correct option: C
-1
Option C (-1) is correct because after substituting z=cos⁡θ±isin⁡θz = \cos \theta \pm i \sin \theta into the expression and simplifying, the result is -1.

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