(1-根号3i)^10-(1 根号3i)^6 (-1 i)^12
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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/(
您好:(-3)^0-根号27+I1-根号2I+根号3+根号2分之1=1-3√3+√2-1+√3+1/2√2=3/2√2-2√3如果本题有什么不明白可以追问,如果满意请点击右下角“采纳为满意回答”如果有
解析:(1-根号3i)^10=2^10×(1/2-根号3/2×i)^10=2^10×[cos(-π/3)+sin(-π/3)×i]^10=2^10×[cos(-10π/3)+sin(-10π/3)×i
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-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
/>i[1-(√3)i]=i×1-i×(√3)i=i+√3
根号32-3倍的根号2分之1+根号18=4倍根号2-(根号2)/6+3倍根号2=7倍根号2-(1/6)根号2=(41/6)倍根号2
(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
不知道楼主什么意思,求近似值吗?=1+1.414+1.732+2+2.236+2.449+2.646+2.828+3+3.162=22.467
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
小弟弟小妹妹,要这样算=(根号1+2+3+4+5+6+7+8+9+10)÷2