(2 2i)^4 (1-根号3i)^5
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/09 10:59:23
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
/>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