TS EAMCET · Maths · Complex Number
\(\omega\) is a complex cube root of unity and Z is a complex number satisfying \(|\mathrm{Z}-1| \leq 2\). The possible values of \(r\) such that \(|\mathrm{Z}-1| \leq 2\) and \(\left|\omega \mathrm{Z}-1-\omega^2\right|=r\) have no common solution are
- A \(0 \leq r \leq 4\)
- B \(r=|\omega|\) only
- C \(r>4\)
- D \(1 < r < 2\)
Answer & Solution
Correct Answer
(C) \(r>4\)
Step-by-step Solution
Detailed explanation
\( |\omega \mathrm{Z}-1-\omega^2|=r \) \( |\omega \mathrm{Z}+\omega|=r \) (since \( 1+\omega+\omega^2=0 \Rightarrow -1-\omega^2=\omega \)) \( |\omega(\mathrm{Z}+1)|=r \) \( |\omega||\mathrm{Z}+1|=r \) \( 1 \cdot |\mathrm{Z}+1|=r \) (since \( |\omega|=1 \)) \( |\mathrm{Z}+1|=r \)…
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