已知复数z满足|z|=1,且z不等于正负i,求证:(z+i)/(z-i)是纯虚数
已知复数z满足|z|=1,且z不等于正负i,求证:(z+i)/(z-i)是纯虚数
数学人气:901 ℃时间:2020-02-05 14:31:44
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令z = x+yi所以x^2 + y^2 = 1(z+i)/(z-i)=[x+(y+1)i]/[x+(y-1)i]=[x+(y+1)i][x-(y-1)i]/[x^2+(y-1)^2]分母是实数 只需证明分子是纯虚数即可分母= x^2-(xy-x)i+(xy+x)i+y^2-1= 2xi 又因为z不等于正负i 所以x不...
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