JEE Advanced 2019 Paper 1 · Mathematics · Complex Numbers · hard

Let ω=e^2π i/7. The real number S=Σk=1^6(1)/(3-ω^k is evaluated. Find S correct to two decimal places.

Numerical answer

Let

\omega=e^{2\pi i/7}

. The real number

S=\displaystyle\sum_{k=1}^{6}\frac{1}{3-\omega^{k}}

is evaluated. Find

S

correct to two decimal places.

Show answer and solution

Answer: 1.83

Let

P(x)=x^7-1=\prod_{k=0}^{6}(x-\omega^k)

. Logarithmic differentiation gives

\frac{P'(x)}{P(x)}=\sum_{k=0}^{6}\frac1{x-\omega^k}=\frac{7x^6}{x^7-1}

. At

x=3

this is

\frac{7\cdot729}{2186}=\frac{5103}{2186}

. The term

k=0

contributes

\frac1{3-1}=\frac12

, which must be removed because the sum in the question starts at

k=1

. Hence

S=\frac{5103}{2186}-\frac12=\frac{5103-1093}{2186}=\frac{4010}{2186}\approx1.8344

. To two decimals the value is

1.83

. The sum is real since the terms for

k

and

7-k

are complex conjugates.

Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake