Rinkoo Gupta
Last Activity: 10 Years ago
{1+cos pi/8 +isin pi/8}/{1+cos pi/8 -isin pi/8}
={2cos^pi/16+i2sinpi/16cospi/16}/2cos^2pi/16-i2sinpi/16cospi/16}
={2cospi/16(cospi/16+isinpi/16)}/{2cospi/16(cospi/16-isinpi/16)}
=e^ipi/16/e^-ipi/16=e^ipi/8
hence solution will be {e^ipi/8)^8=e^ipi=cospi+isinpi=-1
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Rinkoo Gupta
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