Question:
Let \(\alpha\) and \(\beta\) be two roots of the equation \(x^2 + 2x + 2 = 0\), then \(\alpha^{15} + \beta^{15}\) is equal to :
Let \(\alpha\) and \(\beta\) be two roots of the equation \(x^2 + 2x + 2 = 0\), then \(\alpha^{15} + \beta^{15}\) is equal to :
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Find the roots \(-1 \pm i\) and use their squares/powers (\(\alpha^2 = -2i\)) or polar form.
Updated On: Nov 5, 2026
- \(-256\)
- \(512\)
- \(-512\)
- \(256\)
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The Correct Option is A
Solution and Explanation
\(x^2 + 2x + 2 = 0 \Rightarrow x = -1 \pm i\). Take \(\alpha = -1 + i,\ \beta = -1 - i = \bar\alpha\).
Polar form: \(\alpha = \sqrt2\,e^{i\,3\pi/4}\), \(\beta = \sqrt2\,e^{-i\,3\pi/4}\).
\(\alpha^{15} + \beta^{15} = 2(\sqrt2)^{15}\cos\frac{45\pi}{4}\).
\((\sqrt2)^{15} = 2^{7.5} = 128\sqrt2\). \(\frac{45\pi}{4} = 10\pi + \frac{5\pi}{4}\), so \(\cos\frac{45\pi}{4} = \cos\frac{5\pi}{4} = -\frac{1}{\sqrt2}\).
Hence the sum \(= 2 \cdot 128\sqrt2 \cdot \left(-\frac{1}{\sqrt2}\right) = -256\).
Answer: (1) \(-256\)
Polar form: \(\alpha = \sqrt2\,e^{i\,3\pi/4}\), \(\beta = \sqrt2\,e^{-i\,3\pi/4}\).
\(\alpha^{15} + \beta^{15} = 2(\sqrt2)^{15}\cos\frac{45\pi}{4}\).
\((\sqrt2)^{15} = 2^{7.5} = 128\sqrt2\). \(\frac{45\pi}{4} = 10\pi + \frac{5\pi}{4}\), so \(\cos\frac{45\pi}{4} = \cos\frac{5\pi}{4} = -\frac{1}{\sqrt2}\).
Hence the sum \(= 2 \cdot 128\sqrt2 \cdot \left(-\frac{1}{\sqrt2}\right) = -256\).
Answer: (1) \(-256\)
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