InterviewSolution
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InterviewSolution
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If `z=x+iy` is a complex number with `x, y in Q and |z| = 1`, then show that `|z^(2n)-1|` is a rational numberfor every `n in N`. |
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Answer» `|z|=1` `impliesz=e^(itheta)=x+iy` `impliesx=costheta,y=sintheta` Now `costheta" and "sinthetainQ`. Also, `|z^(2n)-1|^(2)=(z^(2n-1)(bar(z)^(2n)-1)` `=(zbar(z))^(2n)-z^(2n)-bar(z)^(2n)+1` `=2-(z^(2n)+bar(z)^(2n))` `=2-2cos2n theta=4sin^(2)ntheta` `implies|z^(2n)-1|=2|sinntheta|` Now, `sinntheta=""^(n)C_(1)cos^(n-1)thetasintheta-""^(n)C_(3)cos^(n-3)thetasin^(3)theta+...` `=" Rational number "(because sintheta, costheta" are rationals")` `implies|z^(2n)-1|=" Rational number "` |
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