UPSC NDA 2 2019 Mathematics Solved Paper

Concept:This problem uses the complex cube roots of unity to simplify large powers.Explanation:Step 1: Recognize the cube roots of unity.$\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.Step 2: Relate these to the given terms.Multiply $\omega$ and $\omega^2$ by $-i$.$\frac{i + \sqrt{3}}{2} = -i\omega$ and $\frac{i - \sqrt{3}}{2} = -i\omega^2$.Step 3: Rewrite the expression $S$.$S = (\frac{i+\sqrt{3}}{2})^{2019} + (\frac{i-\sqrt{3}}{2})^{2019} = (-i\omega)^{2019} + (-i\omega^2)^{2019}$.Step 4: Simplify $(-i\omega)^{2019}$.$(-i)^{2019} = (-i)^{4 \times 504 + 3} = ((-i)^4)^{504} \cdot (-i)^3 = 1 \cdot i = i$.$\omega^{2019} = (\omega^3)^{673} = 1^{673} = 1$.Thus, $(-i\omega)^{2019} = i \cdot 1 = i$.Step 5: Simplify $(-i\omega^2)^{2019}$.$(-i)^{2019} = i$ (as calculated above).$(\omega^2)^{2019} = ((\omega^2)^3)^{673} = 1^{673} = 1$.Thus, $(-i\omega^2)^{2019} = i \cdot 1 = i$.Step 6: Add the results.$S = i + i = 2i$.Answer:$2i$ (Option C).