已知复数z=x+yi(x,y∈R),满足│z│=1,求复数│z-1-i│的取值范围以下是解题过程:依题,由复数z=x+yi(x,y∈R),满足│z│=1,得:x^2+y^2=1另外:│z-1-i│^2=(x-1)^2+(y-1)^2=-2(x+y)+3 (注:将x^2+y^2=1带入)而
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已知复数z=x+yi(x,y∈R),满足│z│=1,求复数│z-1-i│的取值范围以下是解题过程:依题,由复数z=x+yi(x,y∈R),满足│z│=1,得:x^2+y^2=1另外:│z-1-i│^2=(x-1)^2+(y-1)^2=-2(x+y)+3 (注:将x^2+y^2=1带入)而
已知复数z=x+yi(x,y∈R),满足│z│=1,求复数│z-1-i│的取值范围
以下是解题过程:
依题,由复数z=x+yi(x,y∈R),满足│z│=1,得:
x^2+y^2=1
另外:│z-1-i│^2=(x-1)^2+(y-1)^2
=-2(x+y)+3 (注:将x^2+y^2=1带入)
而:1/2=(x^2+y^2)/2 >= [(x+y)/2]^2
所以:(x+y)/2
已知复数z=x+yi(x,y∈R),满足│z│=1,求复数│z-1-i│的取值范围以下是解题过程:依题,由复数z=x+yi(x,y∈R),满足│z│=1,得:x^2+y^2=1另外:│z-1-i│^2=(x-1)^2+(y-1)^2=-2(x+y)+3 (注:将x^2+y^2=1带入)而
3-2√2
=(√2)^2-2×1×√2+1
=(√2-1)^2
3+2√2
=(√2)^2+2×1×√2+1
=(√2+1)^2
依题,由复数z=x+yi(x,y∈R),满足│z│=1,得:
x^2+y^2=1
这里应该是(X+Y)^2=1
|z|=√x^2+y^2=1
x^2+y^2=1
设x=sint y=cost
|z-1-i|=√(x-1)^2+(y-1)^2
=√(sint-1)^2+(cost-1)^2
=√(sin^2t-2sint+1+cos^2t-2cost+1)
=√[-2(sint+cost)+3]
=√[-2√2(sintcos45°+costsin45...
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|z|=√x^2+y^2=1
x^2+y^2=1
设x=sint y=cost
|z-1-i|=√(x-1)^2+(y-1)^2
=√(sint-1)^2+(cost-1)^2
=√(sin^2t-2sint+1+cos^2t-2cost+1)
=√[-2(sint+cost)+3]
=√[-2√2(sintcos45°+costsin45°)+3]
=√[3-2√2sin(t+45°)]
因为-1<=sin(t+45°)<=1
所以√(3-2√2)<=√[3-2√2sin(t+45°)]<=√(3+2√2)
即√(3-2√2)<=|z-1-i|<=√(3+2√2)
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