|z1|=|z2|=1,切z1+z2=3/5+4/5i,求证:z1^2+z2^2=-z1z2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/11/05 16:26:22
|z1|=|z2|=1,切z1+z2=3/5+4/5i,求证:z1^2+z2^2=-z1z2
设Z1=cosa+isina
Z2=cosb+isinb
a∈[0,2π],b∈[0,2π]
欲证Z1^2+Z2^2=-Z1*Z2
即证,-7/25+i24/25-2Z1*Z2=-Z1*Z2
Z1*Z2=-7/25+(24/25)i
Z1+Z2=3/5+i4/5=cosa+cosb+i(sina+sinb)
cosa+cosb=3/5.1)
sina+sinb=4/5.2)
Z1*Z2=cosacosb-sinasinb+i(sinbcosa+sinacosb)
=cos(a+b)+isin(a+b)
Z1^2+Z2^2=cos^2a-sin^2a+cos^2b-sin^2b+i(sin2a+sin2b)
=cos2a+cos2b+i(sin2a+sin2b)
=2cos(a+b)cos(a-b)+2isin(a+b)cos(a-b)
1)^2+2)^2得:
2+2cosacosb+2sinasinb=2+cos(a-b)=1,cos(a-b)=-1/2
所以,Z1^2+Z2^2
=2*[cos(a+b)]*(-1/2)+i[sin(a+b)]*2*(-1/2)
=-cos(a+b)-isin(a+b)
=-Z1*Z2
Z2=cosb+isinb
a∈[0,2π],b∈[0,2π]
欲证Z1^2+Z2^2=-Z1*Z2
即证,-7/25+i24/25-2Z1*Z2=-Z1*Z2
Z1*Z2=-7/25+(24/25)i
Z1+Z2=3/5+i4/5=cosa+cosb+i(sina+sinb)
cosa+cosb=3/5.1)
sina+sinb=4/5.2)
Z1*Z2=cosacosb-sinasinb+i(sinbcosa+sinacosb)
=cos(a+b)+isin(a+b)
Z1^2+Z2^2=cos^2a-sin^2a+cos^2b-sin^2b+i(sin2a+sin2b)
=cos2a+cos2b+i(sin2a+sin2b)
=2cos(a+b)cos(a-b)+2isin(a+b)cos(a-b)
1)^2+2)^2得:
2+2cosacosb+2sinasinb=2+cos(a-b)=1,cos(a-b)=-1/2
所以,Z1^2+Z2^2
=2*[cos(a+b)]*(-1/2)+i[sin(a+b)]*2*(-1/2)
=-cos(a+b)-isin(a+b)
=-Z1*Z2
|z1|=|z2|=1,切z1+z2=3/5+4/5i,求证:z1^2+z2^2=-z1z2
已知复数Z1Z2满足Z1+Z2=2i且|Z1|=|Z2|=|Z1+Z2|,求Z1,Z2
已知z1+z2=3-5i,z1-z2=-1-i,求z1^2+z2^2
已知复数z1z2满足|z1|=|z2|=1z1+z2=-i,求z1.z2
已知复数z1,z2满足z1z2+2i(z1-z2)+1=0,且|z1|=√3,求|z2-4i|
设复数Z1,Z2,满足Z1Z2+2iZ1-2iZ2+1=O .若z1,z2满足z2共轭-z1=2i,求z1,z2
已知复数z1,z2满足|z1|=|z2|=|z1+z2|,且z1+z2=2i,求z1,z2
复数z1,z2满足z1z2≠0,|z1+z2|=|z1-z2|,证明(z1)^2/(z2)^2
已知z1=2-5i,z2=-4+3i,则arg(z1+z2)=
已知复数z1、z2,|z1|=2,|z2|=5,|z1+z2|=6,则|z1-z2|=?
已知z1=-2+i,z1z2=-5+5i,求z1+z2复数的计算
复数 | Z1 |=1, | Z2 |=2, | Z1-Z2 |=根号3,求| Z1+Z2 |