A) 512
B) \[-\,512\]
C) \[-\,256\]
D) 256
Correct Answer: C
Solution :
| \[\alpha ,\,\beta \] are roots of \[{{x}^{2}}+2x+=\,0\] |
| \[\Rightarrow \] \[(x+1){{\,}^{2}}=\,-1\,\]\[\Rightarrow \] \[x\,=\,-\,1\pm \,i\] |
| Let \[\alpha \,=\,-1\,+\,i,\] \[\beta \,=\,-1-i\] |
| \[{{\alpha }^{15}}+{{\beta }^{15}}\,=\,{{(-1+i)}^{15}}+\,{{(-1\,-i)}^{15}}\] |
| \[=\,-\,\left[ \,{{(1+\,i)}^{15}}+\,{{(1-i)}^{15}} \right]\] |
| \[=\,\,-\left[ {{\left\{ \sqrt{2}\left( \cos \frac{\pi }{4}+\,i\,\sin \,\frac{\pi }{4} \right) \right\}}^{15}}+{{\left\{ \sqrt{2}\left( \cos \left( -\,\frac{\pi }{4} \right)+\,i\,\sin \,\left( -\frac{\pi }{4} \right) \right) \right\}}^{15}} \right]\,\] |
| \[=\,\,-\,\left[ {{\left( \sqrt{2} \right)}^{15}}\left\{ \text{cos}\frac{15\pi }{4}+\text{isin}\frac{15\pi }{4} \right\}+{{(\sqrt{2})}^{15}}\left\{ \text{cos}\frac{15\pi }{4}-\text{i}\,\text{sin}\frac{15\pi }{4} \right\} \right]\] |
| \[=\,\,-\,{{(\sqrt{2})}^{15}}\left[ 2\text{cos}\frac{15\pi }{4} \right]\,=-\,256.\] |
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