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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Complex Number

NDA 2023 · Mathematics · Complex Number

If α and β are the distinct roots of equation x^2-x+1=0 , then what...

If αand\alpha and\betaarethedistinctrootsofequationare the distinct roots of equationx^2-x+1=0,thenwhatisthevalueof∣α100+β100α100−β100∣, then what is the value of \left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right| ?

Options

  1. A3\sqrt{3}
  2. B2\sqrt{2}
  3. C11
  4. D13\frac{1}{\sqrt{3}}

Correct answer

D. 13\frac{1}{\sqrt{3}}

Step-by-step solution

x2−x+1=0x^2-x+1=0 We know that roots of x2+x+1=0x^2+x+1=0 are ω\omega and ω2\omega^2. ∴\therefore \quad Roots of x2−x+1=0x^2-x+1=0 are −ω-\omega and −ω2-\omega^2 where ω,ω2\omega, \omega^2 are complex cube roots of unity ∴∣α100+β100α100−β100∣=∣ω100+(ω2)100ω100−(ω2)100∣=∣ω+ω2ω−ω2∣=∣−1ω−ω2∣=∣−1−1+3i2−(−1−3i)2∣=∣−13i∣=∣i3∣=13\therefore\left|\frac{\alpha^{100}+\beta^{100}}{\alpha^{100}-\beta^{100}}\right|=\left|\frac{\omega^{100}+\left(\omega^2\right)^{100}}{\omega^{100}-\left(\omega^2\right)^{100}}\right|=\left|\frac{\omega+\omega^2}{\omega-\omega^2}\right|=\left|\frac{-1}{\omega-\omega^2}\right|=\left|\frac{-1}{\frac{-1+\sqrt{3} i}{2}-\frac{(-1-\sqrt{3} i)}{2}}\right|=\left|\frac{-1}{\sqrt{3} i}\right|=\left|\frac{i}{\sqrt{3}}\right|=\frac{1}{\sqrt{3}}
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