(2 2i)^4 (1-根号3i)^5-搜答案-拍题作业网

原式=(1-√3i)/(√3+i)^2=(1-√3i)/(2+2√3i)=(1-√3i)^2/[2*(1-3)]=(-2-2√3i)/(-4)=(1+√3i)/2.若前面没有括号,原式=1-√3i/(

因为：（1+i)/(1-i)=(1+i)²/(1-i)(1+i)=2i/2=i,则[（1+i)/（1-i)]^6=i^6=(i²)³=(-1)³=-1.又因为：

【(1-根号3*i)/(1+i)】^2＝［（1－根号3i)(1-i)/(1+1)]^2=[(1-i-根号3i-根号3）／2］^2=[(1-根号3）－（1＋根号3）i]^2/4=［(1-根号3）^2-2

(2+2i)/(3^.5-i)=(2+2i)(3^.5+i)/[3+1]=0.5(1+i)(3^.5+i)(2-2i)/(1+3^.5i)=(2-2i)(1-3^.5i)/[1+3]=0.5(1-i)

(2+2i)/(√3-i)=(2+2i)(√3+i)/(3+1)=0.5(1+i)(√3+i)(2-2i)/(1+√3i)=(2-2i)(1-√3i)/(1+3)=0.5(1-i)(1-√3i)=-0

[（2+2i)/(√3-i)]^7=[(1+i)(√3+i)/2]^7=(√2)^7*[(√2/2+√2/2i)(√3/2+1/2i)]^7=(√2)^7*[(cosπ/4+isinπ/4)(cosπ

i-2√3/(1+2√3i)+(5+i^19)-(1+i/√2)^2=i-(2√3)(1-2√3i)/13+5-i-(1+i-1/4)=i-(2√3/13-12i/13)+5-i-3/4-i=17/4

(√3+√3i)³·(3-4i)/4+3i=3√3(1+i)³·(3-4i)/4+3i=3√3·2i(1+i)(3-4i)/4+3i=3√3·2i(7-i)/4+3i=(21√3i

|√3+i|=2=>|√3+i|^4=2^4|2-2i|=2√2=>|2-2i|^4=2^6|1-√3i|=2=>|1-√3i|^8=2^8∴|z|=2^(4+6-8)=4

进行分母有理化,上下同时乘以(1+根号3i)^5

计算i(1-根号3i

/>i[1-(√3)i]=i×1-i×(√3)i=i+√3

[(-1+根号3i)^3]/[(1+i)^6]+[-2+i]/[1+2i]=[(-1+根号3i)³]/(2i)³+[(-2+i)(1-2i)]/[(1+2i)(1-2i)]=[(-

（2根号3i+6i）/3

(√3-i)/(1+√3i)=(√3-i)(1-√3i)/(1+√3i)(1-√3i)=(√3-√3-i-3i)/(1+3)=-i(1+根号3i)/(根号3-i）刚好为上面的倒数因此=1/(-i)=i

逐项进行计算-2根号3+i/(1+2根号3i）=(-2√3+i)(1-2√3i)/13=i（根号2/（1+i））^2010=((1-i)/√2)^2010=(-i)^1005=-i^1005=-i所以

(-1+sqrt(3)i)^4/(1-i)^8=1/16(-1+isqrt(3))^4=1/2(-1+isqrt(3))为x^3-1=0的根(-1+sqrt(3i))^4/(1-i)^8=-1/2+s

=(24+7i)/32

答案是：i将分子分母同时乘以根号3+i,然后分母就变成了3-i^2=4,而分子变成了4i,这样结果就是i