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發表於 2005-3-17 07:08:52
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顯示全部樓層
a.i) z^2n=-1,
cos(pi) + isin(pi)=-1,
hence cos(pi) + isin(pi) is a root of z^2n, so as cos(pi +2kpi) + isin(pi +2kpi), for k=0,1,..........
to find the roots of z, apply the De Moivre's Them.
z^2n = cos(pi + 2kpi) + isin(pi +2kpi)
z=cos((pi +2kpi)/2n) + isin((pi+2kpi)/2n),
the power 2n indicates there are 2n roots in total, so k takes value of 0,1,...2n-1
a.ii) first note that the roots of any z occurs in conjugate pair,
first of the solution just proof that in details,
then since z1, z2,.. zk, are roots of z, muitiple of (z-z1)=(z-z2)=(z-zk)=0
mutiple of all (z-zk) still=0=LHS,
and then couple more of maths skills will solve the prob..
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