JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(z = a +i b , b \neq 0\) be complex numbers satisfying \(z ^{2}=\overline{ Z } \cdot 2^{1-|z|}\). Then the least value of \(n\) \(\in N\), such that \(z ^{ n }=( z +1)^{ n }\), is equal to.
- A \(0\)
- B \(6\)
- C \(5\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
\(\left|z^{2}\right|=|\bar{z}| \cdot 2^{1-|z|} \Rightarrow|z|=1\) \(z ^{2}=\overline{ z } \Rightarrow z ^{3}=1 \therefore z =\omega\) or \(\omega^{2}\) \(\omega^{ n }=(1+\omega)^{ n }=\left(-\omega^{2}\right)^{ n }\) Least natural value of \(n\) is \(6.\)
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