若數列{An}的各項依次是關于x的方程(1-i)^x=i(1+i)^x（i是虛數單位）的從小到大的全部正實數解，那么lim{(n+1)/(An-4n)}=_____

熱心網友

最怕復數∵(1-i)^x=i(1+i)^x∴i=[(1-i)^x]/[(1+i)^x]=[(1-i)/(1+i)]^x=[(1-i)^2/(1-i^2)]^x=(-i)^x∴x=3;7;11;15　………… an=3+4(n-1)=4n-1lim{(n+1)/(An-4n)}=lim{(n+1)/[(4n-1)-4n]}=lim{(n+1)/(-1)}=-n或-∞可這不是極限啊

熱心網友

解:∵i[(1+i)^x]=(1-i)^x ∴i=[(1-i)/(1+i)]^x=[(1-i)(1-i)/2]^x=(-i)^x即:i=(-i)^x得i*(-i)=(-i)^x*(-i)(-i)^(x+1)=1=(-i)^4n(x+1)=4nx=4n-1(n∈N+)

熱心網友

解:原方程可化為(1-i/1+i)的x次方=i,即(-i)的x次方=i,方程的正實數解為4n-1,An-4n=-1,所求極限=Lim(n+1)/- 1=-∞是算錯了,還是題目數椐有誤?

熱心網友

(1+i)^x=(1-i)^x---[(1+i)/(1-i)]^x=1---(2i/2)^x=i^x=1.i^(4n)=1。見中學教材：i^(4n）=(cos90+isin90)^(4n)=cos(n*360)+isin(n*360)=1---x=4n(n∈N)---An=4n.---An-4n=0???!!!if An-8n---lim(n+1)/(4n-8n)=lim(1+1/n)/(-4=(1+0/(-4)=-1/4

熱心網友

剛才我看錯了(1-i)^x=[(√2)(cos(π/4)-isin(π/4))]^x=2^(x/2)*(cos(xπ/4)-isin(xπ/4))i(1+i)^x=[(√2)(cos(π/4)+isin(π/4))]^x=i*2^(x/2)*(cos(xπ/4)+isin(xπ/4))=2^(x/2)*(-sin(xπ/4)+icos(xπ/4))所以cos(xπ/4)=-sin(xπ/4)-sin(xπ/4)=cos(xπ/4)sin(xπ/4)+cos(xπ/4)=√2(sin((x+1)π/4)=0x=4n-1所以An=4n-1lim{(n+1)/(An-4n)}=lim[(n+1)/(-1)]=-n-1

熱心網友

A=4?????