(1 2X 2 3Y-3 4Z)-(1 2X-2 3Y-3 4Z)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 14:15:39
(1 2X 2 3Y-3 4Z)-(1 2X-2 3Y-3 4Z)
已知复数z满足3z+(z-2)i=2z-(1+z)i,求z

设z=a+bi因为3z+(z-2)i=2z-(1+z)i所以3(a+bi)+(a+bi-2)i=2(a+bi)-(1+a+bi)i3a+3bi+ai-b-2i=2a+2bi-i-ai+b(3a-b)+

已知复数z满足z*z-3i*z=1+3i,求z

z*z-3i*z=1+3i化简(z+1)(z-1-3i)=0所以z=-1或z=1+3i

复数Z满足1/Z=Z/(3Z-10)则,|Z|=

1/Z=Z/(3Z-10)即:z²=3z-10z²-3z+10=0∴z=(3±i*√31)/2|Z|=√10

已知模(z+1)/z=2 arg[(z+1)/z]=π/3 求z.

则由题意得,(z+1)/z=2(cosπ/3+sinπ/3*i),设z=a+bi(a+bi+1)/a+bi=2(cosπ/3+sinπ/3*i)a+1+bi=(a-sqrt(3))+(sqrt(3)a

虚数Z满足Z的模=1,Z^2+2Z+1/Z

虚数z满足|z|=1,z²+2z+1/z

已知复数z满足z+1/z∈R,|z-2|=2,求z

设z=a+bi,a,b是实数|z-2|^2=(a-2)^2+b^2=41/z=1/(a+bi)=(a-bi)/(a^2-b^2)z+1/z=[a+a/(a^2-b^2)]+[b-b/(a^2-b^2)

已知Z-|Z|=-1+i,求复数Z

设z=a+bi代入得a+bi-√(a^2+b^2)=-1+i比较两边得a-√(a^2+b^2)=-1b=1代入得a-√(a^2+1)=-1-√(a^2+1)=-1-a平方得a^2+1=a^2+2a+1

解二元一次方程组X:Y:Z=1:2:3,X+Y+Z=12

X:Y:Z=1:2:3则Y=2X,Z=3X代入X+Y+Z=12,得X+2X+3X=12解得X=2,则Y=2X=4Z=3X=6

已知{x:y:z=1:2:3,x+y+z=12,求x、y、z的值

x:y:z=1:2:3,x=k,y=2k,z=3kx+y+z=k+2k+3k=6k=12k=2x=2,y=4,z=6

已知x、y、z满足|4x-4y+1|+152y+z+(z−12)

根据题意得,4x-4y+1=0,2y+z=0,z-12=0,解得x=-12,y=-14,z=12,∴x+z-y=-12+12-(-14)=14,∴x+z−y=14=12.故答案为:12.

已知:f(z)=|1+Z|-.Z

f(Z)=|1+z|-.Z,f(-z)=|1-z|+.Z设z=a+bi  (a、b∈R)  由f(-z)=10+3i得|1-(a+bi)|+a-bi=10+3i

x:y=1:2 z:x=3:1 x +y+z=12

由x:y=1:2,得y=2x,由z:x=3:1,得z=3x,代人x+y+z=12中,得,x+2x+3x=12,解得,x=2,所以y=2x=4,z=3x=6

已知模[(z+1)/z]=2 arg[(z+1)/z]=π/3 求z.

因为模[(z+1)/z]=2arg[(z+1)/z]=π/3所以(z+1)/z=2(cosπ/3+isinπ/3)1+1/z=1+√3i1/z=√3iz=1/[√3i]=-√3/3i

把F(z)=1/z(z-1)在1

点击放大:

1.x+y+z=21,x-y=1,2x+z-y=13.2.3x+2y+z=13,x+y+2z=7 ,2z+3y-z=12

1.x=10,y=9,z=22.x=3,y=2,z=13.x=30,y=20,z=16.

复数z满足(z-1)(2-z)=5

复数z满足(z-1)(2-z)=52z-2-z^2+z=5这里z²;相当于i²=-1则3z=5+2-1=63z=6z=2

虚数z满足绝对值z=1,且z^2+2z+1/z

z=cost+isintcos2t+isin2t+2cost+2isint+cost-isint

若复数z满足,z*z拔+(1-2i)*z+(1+2i)z拔

设z=a+bi,则:z拔=a-bi.则:z*z拔=(a+bi)(a-bi)=a²+b²(1-2i)z+(1+2i)z拔=(z+z拔)+2i(z拔-z)=2a+4b则:a²