Question

复变函数基础题 (1+i)^100+(1-i)^100 答案是2^51 求指数形式解题过程

Answer 1

这题关键是利用欧拉公式，e^(iθ)=cosθ+isinθ
1+i=(根号2)*[(cos(π/4)+isin(π/4)]=(根号2)*e^(i*π/4)
(1+i)^100=(根号2)^100*e(i*25π)=2^50*(cos25π+isin25π)
=-2^50
同理(1-i)^100=-2^50
(1+i)^100+(1-i)^100=-2^50-2^50=-2^51
答案为什么是正的2^51 呢，是不是看错了

Answer 2

1+i = √2(sin(π/4) + icos(π/4))
(1+i)^100= 2^(50)[sin(25π)-cos(25π)] =2^50
1-i = √2(sin(-π/4) + icos(-π/4))
(1-i)^100= 2^50[sin(-25π)-cos(-25π)] = 2^50
(1+i)^100+(1-i)^100 =2^51

Answer 3

由1+i=(√2)e^(iπ/4),1-i=(√2)e^(-iπ/4)
那么(1+i)^100+(1-i)^100=(2^50)[e^(25πi)+e^(-25πi)]=2^51

Answer 4

(1+i)^100+(1-i)^100
=[(1+i)^2]^50+[(1-i)^2]^50
=(2i)^50+(-2i)^50
=(2^50)(i^50)+(2^50)(-i)^50
=2^50+2^50
=2*2^50
=2^51