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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Complex Number

NDA 2024 · Mathematics · Complex Number

If ω ≠ 1 is a cube root of unity, then what are the solution of...

If ω≠1\omega \neq 1 is a cube root of unity, then what are the solution of (z−100)3+1000=0(z-100)^3+1000=0 ?

Options

  1. A10(1−ω),10(10−ω2),10010(1-\omega), 10\left(10-\omega^2\right), 100
  2. B10(10−ω),10(10−ω2),9010(10-\omega), 10\left(10-\omega^2\right), 90
  3. C10(1−ω),10(10−ω2),100010(1-\omega), 10\left(10-\omega^2\right), 1000
  4. D(1+ω),(10+ω2),−1(1+\omega),\left(10+\omega^2\right),-1

Correct answer

B. 10(10−ω),10(10−ω2),9010(10-\omega), 10\left(10-\omega^2\right), 90

Step-by-step solution

(z−100)3+1000=0⇒(z−100)3=−1000=(−10)3⇒z−100=−10⇒z=90z−100=10w⇒z=100−10wz−100=−10w2⇒z=100−10w2.\begin{aligned} & (z-100)^3+1000=0 \\ & \Rightarrow \quad(z-100)^3=-1000=(-10)^3 \\ & \Rightarrow \quad z-100=-10 \\ & \Rightarrow \quad z=90 \\ & z-100=10 w \Rightarrow z=100-10 w \\ & z-100=-10 w^2 \\ & \Rightarrow \quad z=100-10 w^2. \end{aligned}
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