[(1+i)/(1-i)]的2005次幂等于多少?

[(1+i)/(1-i)]的2005次幂等于多少?
[(1+i)/(1-i)]的2005次幂等于多少?

[(1+i)/(1-i)]的2005次幂等于多少?
[(1+i)/(1-i)]的2005次幂
=[(1+i)²/（1+i）（1-i）]的2005次幂
=[（1+2i-1）/(1+1)]的2005次幂
=i的2005次幂
=-i

先算[(1+i)/(1-i)]
[(1+i)/(1-i)]
=[(1+i)²/（1+i）（1-i）]
=[（1+2i-1）/(1+1)]
=i
i的2005次幂=i

(1＋i)/(1－i)
=[(1＋i)²]/[(1－i)(1＋i)]
=(1＋2i－1)/(1＋1)
=i
则：
[(1＋i)/(1－i)]的2005次方
=i的2005次方 【i的2004次方=1】
=i

∵　[(1＋i)/(1－i)] = [[(1＋i)×(1+i)]/[(1＋i)×(1－i)]
= (1＋i)2/[12－(－i)2]
=(12+ 2i + i 2 )/( 1 －i2)
= 2i/2
= i
∴[(1＋i)/(1－i)] 2005
= i2005
= i 2004×i
=（i2)1002×i
= i

[(1+i)/(1-i)]^2005 分子分母同时×（1+i　）得
[(1+i)^2/(1+i)(1-i)^2]^2005
=[1^2+i^2+2i/(1-i^2)]^2005
=(2i/2)^2005
=i^2005
=i