JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(z = \frac{{\sqrt 3 }}{2} + \frac{i}{2}\,\,\,\left( {i = \sqrt { - 1} } \right)\), then \({\left( {1 + iz + {z^5} + i{z^8}} \right)^9}\) is equal to
- A \(-1\)
- B \(1\)
- C \((-1 + 2i)^9\)
- D \(0\)
Answer & Solution
Correct Answer
(A) \(-1\)
Step-by-step Solution
Detailed explanation
\(z=\frac{\sqrt{3}+i}{2}=e^{\frac{i \pi}{6}}\) \(\left(1+i z+z^{5}+i z^{8}\right)^{9}\) \(=\left(1+\mathrm{e}^{\mathrm{i} \pi / 2} \mathrm{e}^{\mathrm{i} \pi / 6}+\mathrm{e}^{\mathrm{i} 5 \pi / 6}+\mathrm{e}^{\mathrm{i} \pi / 2} \mathrm{e}^{\mathrm{i} 8 \pi / 6}\right)^{9}\)…
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