JEE Mains · Maths · STD 11 - 4.1 complex nubers
The value of \(\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3\) is
- A \(\frac{-1}{2}(1-i \sqrt{3})\)
- B \(\frac{1}{2}(1-i \sqrt{3})\)
- C \(\frac{-1}{2}(\sqrt{3}-i)\)
- D \(\frac{1}{2}(\sqrt{3}+i)\)
Answer & Solution
Correct Answer
(C) \(\frac{-1}{2}(\sqrt{3}-i)\)
Step-by-step Solution
Detailed explanation
Let \(\sin \frac{2 \pi}{9}+ i \cos \frac{2 \pi}{9}= z\) \(\left(\frac{1+ z }{1+\bar{z}}\right)^3=\left(\frac{1+ z }{1+\frac{1}{ z }}\right)^3= z ^3\) \(\Rightarrow\left( i \left(\cos \frac{2 \pi}{9}- i \sin \frac{2 \pi}{9}\right)\right)^3\)…
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