1.(1-i/1+i)^1002.求商：3+2i/5+7i3.(√3+i)^5/(-1+√3i)
热心网友
1.[(1-i)/(1+i)]^100=[(1-i)^2/(1+i)(1-i)]^100=[(1-2i-1)/2]^100=(-i)^100=12.3+2i/5+7i=(3+2i)(5-7i)/(5+7i)(5-7i)=(29-11i)/743.(√3+i)^5/(-1+√3i)=(√3+i)^4*(3+i)(-1-√3i)/(-1+√3i)(-1-√3i)=(3+2√3i-1)^2*(-4i)/4=-i(-8+8√3i)=8√3+8i
热心网友
1.[(1-i)/(1+i)]^100==[(√2/2-√2i/2)/((√2/2+√2i/2)]^100==e^[(-πi/4-πi/4)100]=e^(-50πi)=12.(3+2i)/(5+7i)=(3+2i)(5-7i)/74=(29-11i)/74.3.(√3+i)^5/(-1+√3i)=2^4(√3/2+i/2)^5/(-1/2+√3i/2)==2^4e^(5πi/6-2πi/3)=2^4e^(πi/6)=2^4(√3/2+i/2)=8(√3+i).