设n是奇数,则((1+i)/√2)^4n+((1-i)/√2)^4n=?
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设n是奇数,则((1+i)/√2)^4n+((1-i)/√2)^4n=?
由欧拉定理,
(1+i)/√2=cos(PI/4) + i*sin(PI/4) = e^[iPI/4],
((1+i)/√2)^(4n) = e^[iPI/4*4n] = e^[inPI] = cos(nPI) + i*sin(nPI) = cos(nPI) = (-1)^n
(1-i)/√2=cos(-PI/4) + i*sin(-PI/4) = e^[-iPI/4],
((1-i)/√2)^(4n) = e^[-iPI/4*4n] = e^[-inPI] = cos(-nPI) + i*sin(-nPI) = cos(nPI) = (-1)^n
n是奇数.
((1+i)/√2)^(4n)+((1-i)/√2)^(4n) = 2(-1)^n = -2.
(1+i)/√2=cos(PI/4) + i*sin(PI/4) = e^[iPI/4],
((1+i)/√2)^(4n) = e^[iPI/4*4n] = e^[inPI] = cos(nPI) + i*sin(nPI) = cos(nPI) = (-1)^n
(1-i)/√2=cos(-PI/4) + i*sin(-PI/4) = e^[-iPI/4],
((1-i)/√2)^(4n) = e^[-iPI/4*4n] = e^[-inPI] = cos(-nPI) + i*sin(-nPI) = cos(nPI) = (-1)^n
n是奇数.
((1+i)/√2)^(4n)+((1-i)/√2)^(4n) = 2(-1)^n = -2.
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