A) -256
B) 512
C) -512
D) 256
Correct Answer: A
Solution :
Equation \[{{x}^{2}}+2x+2=0\] has roots \[\alpha \], \[\beta \] \[{{\left( x+1 \right)}^{2}}+1=0\] \[\,x=\,\,-1=\pm \,\,i\] \[x=-1\text{ }\pm \text{ }i\] Let \[\alpha \,=\,\,-1\,\,+\,\,i\] \[\beta \,\,=\,\,-1\,\,+\,\,i\] \[\alpha \,\,=\,\,\sqrt{2}{{e}^{i3\pi /4}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\beta \,\,=\,\,\sqrt{2}\,{{e}^{i5\pi /4}}\] \[\therefore \,\,\,{{a}^{15}}\,\,+\,\,{{\beta }^{15}}\,\,=\,{{(\sqrt{2}{{e}^{i3\pi /4}})}^{15}}\,+\,\,{{(\sqrt{2}\,{{e}^{i5\pi /4}})}^{15}}\] \[=\,{{(\sqrt{2})}^{15}}\,\left[ \cos \,\frac{45\pi }{4}+i\sin \,\frac{45\pi }{4}+\cos \,\frac{75\pi }{4}+i\sin \,\frac{75\pi }{4} \right]\] \[=\,\,\,{{2}^{15/2}}\,\left[ \frac{-1}{\sqrt{2}}-\frac{i}{\sqrt{2}}-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \right]\] \[=\text{ }-2\]8 = -256
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