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z=3+3i,或z=-2-2i.
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+b)a+1=a-sqrt()3b=sq
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
虚数z满足|z|=1,z²+2z+1/z
z=cost+isintcos2t+isin2t+2cost+2isint+cost-isint
1/Z=Z/(3Z-10)即:z²=3z-10z²-3z+10=0∴z=(3±i*√31)/2|Z|=√10
z*z-3i*z=1+3i化简(z+1)(z-1-3i)=0所以z=-1或z=1+3i
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)]i∈R所以b-b/(a^2-b^2)=
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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+1a=0因此z=i
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即(1−a)2+b2+a=10−b=3,
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
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)+(3b+a-2)i=(2a+b)+(2b
设z=a+bi(a、b为实数,且b≠0)(1-z)/(1+z)=i1-z=(1+z)i1-a-bi=(1+a+bi)i整理,得(a-b-1)+(a+b+1)i=0a-b-1=0a+b+1=0解得a=0b=-1z=-iz+1=1-i|z+1|
1、析:设z=a+bi,b≠0则z+1/z=a+bi+1/(a+bi)=a+bi+(a-bi)/(a^2+b^2)=(a+a/(a^2+b^2)+[b-b/(a^2+b^2]i,∴b-b/(a^2+b^2=0,解得b=0(舍去)或a^2+b
由|z|=1设z=cosθ+isinθ(θ∈[0,2π)由z^5+z=1得cos5θ+cosθ+i(sin5θ+sinθ)=1于是cos5θ+cosθ=1,sin5θ+sinθ=0由sin5θ+sinθ=0得θ=0,π/3,2π/3,π,4
设z=x+yi(x,y为实数)1=|z+1|^2-|z-i|^2=|(x+1)+yi|^2-|x+(y-1)i|^2=(x+1)^2+y^2-[x^2+(y-1)^2]=x^2+2x+1+y^2-(x^2+y^2-2y+1)=2x+2y即:
设z=yi原式=yi/1+y——i²=-1
设z=a+bi,1/(a+bi)=(a-bi)/(a^2+b^2)=1/2,显然b=0,a/(a^2+b^2)=1/2;a=2.得z=2