复数1 根号3i 根号3-i

针对你的题目,将复数写成指数形式算是较为简便的：exp(iπ/4)=cos(π/4)+isin(π/4)=(1+i)/√2exp(-iπ/6)=(1-i√3)/2所以原式=[(2√2)^4]exp(i

μ=（根号3/2+1/2i）/(-1+根号3i)|μ|=|（根号3/2+1/2i）/(-1+根号3i)|因为|根号3/2+1/2i|=1|-1+根号3i|=2所以|μ|=0.5

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

根号3?(1/2+√3/2i)^3=(cos(π/3)+isin(π/3))^3=cosπ+isinπ=-1

2i/(-1+√3i)=[2i*(√3i+1)]/[(√3i+1)(√3i-1)]=(2i-2√3)/(-4)=(√3/2)-(i/2)所以虚部为-i/2实部为√3/2

=(1+2*3^0.5*i-3)*(1+3^0.5*i)=(-2+2*3^0.5*i)*(1+3^0.5*i)=-2*(1-3^0.5*i)*(1+3^0.5*i)=-2*(1+3)=-8=[2*(c

设z=a+bi|a+bi+√3+i|=|(a+√3)+(b+1)i|=√[(a+√3)²+(b+1)²]=1|(a+√3)²+(b+1)²=1令a=-√3+si

欧拉规定：i=根号-1,顺便说一句,i应该叫虚数.楼上也太懒了~

答：把原式化简成三角形式.利用[r(cosθ+isinθ)]^n=（r^n）[cosnθ+isinnθ]公式计算[cos（π/3）+sin(π/3)i]^5+[cos(2/3π）+sin(2/3π)i

|√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

Z=(√3-i)/(1-3+2√3i)=(√3-i)/2(√3i-1)=(√3-i)(√3i+1)/2(-3-1)=(3i+√3+√3-i)/(-8)=-√3/4-i/4;∴a=-√3/4;b=-1/

原式=[2(cos2π/3+isin2π/3)]^5/[2(cosπ/3+isinπ/3)=32(cos10π/3+usin10/3)/[2(cosπ/3+isinπ/3)]=16(cos3π+isi

(√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

(-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

z=(根号3-i)/[1+i根号3]=(根号3-i)*[1-i根号3]/{[1-i根号3][1+i根号3]}=(-4i)/(1+3)=-iz的模为1

用角度制√3i-1=2(cos30-i*sin30)三次根号下[（根号3）i-1]=三次根号下2*(cos10-i*sin10)-√3i-1=2(-cos30-i*sin30)三次根号下[-（根号3）

=(24+7i)/32

Z=(1-i)/(√3/2+1/2i)=(1-i)(√3/2-1/2i)/(3/4+1/4)=(1-i)(√3/2-1/2i)=√3/2-1/2i-√3/2i+1/2i²=(√3/2-1/2

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

设z=a+bi可得：(1+i)(a+bi)=a+ai+bi+bi^2=(a-b)+(a+b)i=1+√3i所以可得：a-b=1a+b=√3解得：a=(√3+1)/2,b=(√3-1)/2|z|=√(a