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JEE Advanced 2025 Paper 2 · Mathematics · Complex Numbers · hard

Let zeta=e^2π i/7. The value of Σk=1^6(1)/(3-zeta^k is ______.

Numerical answer

Let ζ=e2πi/7\zeta=e^{2\pi i/7}. The value of ∑k=1613−ζk\displaystyle\sum_{k=1}^{6}\frac{1}{3-\zeta^k} is ______. Give the answer correct to two decimal places.
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Answer: 1.83

The numbers ζk\zeta^k, k=1,…,6k=1,\dots,6, are the roots of P(x)=x7−1x−1=1+x+⋯+x6P(x)=\frac{x^7-1}{x-1}=1+x+\dots+x^6. So P(x)=∏k=16(x−ζk)P(x)=\prod_{k=1}^6(x-\zeta^k) and P′(x)P(x)=∑k=161x−ζk\dfrac{P'(x)}{P(x)}=\sum_{k=1}^{6}\dfrac{1}{x-\zeta^k}. At x=3x=3: P(3)=37−12=1093P(3)=\frac{3^7-1}{2}=1093, and P′(3)=1+2⋅3+3⋅9+4⋅27+5⋅81+6⋅243=2005P'(3)=1+2\cdot3+3\cdot9+4\cdot27+5\cdot81+6\cdot243=2005. The sum is therefore 20051093≈1.8344\frac{2005}{1093}\approx1.8344, i.e. 1.831.83. Direct numerical evaluation with complex arithmetic gives the same real value.
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