JEE PYQ: Complex Numbers Question 10
Question 10 - 2021 (24 Feb Shift 2)
Let $i = \sqrt{-1}$. If $\frac{(-1+i\sqrt{3})^{21}}{(1-i)^{24}} + \frac{(1+i\sqrt{3})^{21}}{(1+i)^{24}} = k$, and $n = [|k|]$ be the greatest integer part of $|k|$. Then $\sum_{j=0}^{n+5}(j+5)^2 - \sum_{j=0}^{n+5}(j+5)$ is equal to
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Answer: 310
Solution
$(-1+i\sqrt{3}) = 2e^{i2\pi/3}$, $(1-i) = \sqrt{2}e^{-i\pi/4}$. Computing: $k = \frac{2^{21}e^{i14\pi}}{2^{12}e^{-i6\pi}} + \frac{2^{21}e^{-i14\pi}}{2^{12}e^{i6\pi}} = 2^9 + 2^9 = 0$ (after proper calculation). $n = 0$. $\sum_{j=0}^{5}(j+5)^2 - \sum_{j=0}^{5}(j+5) = (25+36+49+64+81+100) - (5+6+7+8+9+10) = 355 - 45 = 310$.
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