Question:
What is the value of \(\left(\frac{-1+i\sqrt{3}}{2}\right)^{3n} + \left(\frac{-1-i\sqrt{3}}{2}\right)^{3n}\), where \(i = \sqrt{-1}\)?
\(\left(\frac{-1+i\sqrt{3}}{2}\right)^{3n} + \left(\frac{-1-i\sqrt{3}}{2}\right)^{3n}\) का मान क्या है, जहाँ \(i = \sqrt{-1}\)?
Options:
- 3
- 2
- 1
- 0
- 3
- 2
- 1
- 0
Correct Answer: b)
Explanation:
The complex cube roots of unity are:
\[\omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 – i\sqrt{3}}{2}\]
We know that \(\omega^3 = 1\).
Therefore:
\[\omega^{3n} = (\omega^3)^n = 1^n = 1\]
\[(\omega^2)^{3n} = (\omega^3)^{2n} = 1^{2n} = 1\]
Adding the two values:
\[\omega^{3n} + (\omega^2)^{3n} = 1 + 1 = 2\]
Hence, the correct option is (b).
इकाई के काल्पनिक घनमूल:
\[\omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 – i\sqrt{3}}{2}\]
हम जानते हैं कि \(\omega^3 = 1\) होता है।
अतः:
\[\omega^{3n} = (\omega^3)^n = 1^n = 1\]
\[(\omega^2)^{3n} = (\omega^3)^{2n} = 1^{2n} = 1\]
दोनों का योग:
\[1 + 1 = 2\]
अतः सही विकल्प (b) है।