JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(z ^{2}+ z +1=0, z \in C\), then \(\left|\sum_{ n =1}^{15}\left( z ^{ n }+(-1)^{ n } \frac{1}{ z ^{ n }}\right)^{2}\right|\) is equal to
- A \(1\)
- B \(3\)
- C \(2\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(z ^{2}+ z +1=0 \Rightarrow z = w , w ^{2}\) \(\left|\sum_{n=1}^{15}\left(z^{n}+(-1) \frac{1}{z^{n}}\right)^{2}\right|=\left|\sum_{n=1}^{15}\left(z^{2 n}+\frac{1}{z^{2 n}}+2(-1)^{n}\right)\right|\) \(=\left|\sum_{n=1}^{15} w^{2 n}+\frac{1}{w^{2 n}}+2(-1)^{n}\right|\)…
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