|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