(1+i)^2n/1-i+(1-i)^2n/1+i=2^n,求最小正整数n?
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(1+i)^2n/1-i+(1-i)^2n/1+i=2^n,求最小正整数n?
(1+i)^2n/1-i+(1-i)^2n/1+i=2^n,求最小正整数n?
(1+i)^2n/1-i+(1-i)^2n/1+i=2^n,求最小正整数n?
(1+i)^2n/1-i+(1-i)^2n/1+i=2^n
通分(1+i)^(2n+1)+(1-i)^(2n+1)=2^n(1-i²)
1-i=-i²-i=-i(1+i),1-i²=1+1=2
所以(1+i)^(2n+1)-i(1+i)^(2n+1)=2^(n+1)
合并:(1+i)^(2n+1)*(1-i)=2^n(1-i²)
(1+i)^(2n+1)=2^n(1+i)
(1+i)^2n=2^n
(1+i)²^n=2^n
(2i)^n=2^n
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