一道关于复数的题目已知复数Z满足
设z=cost+isint
--->|z|=1,1/z=z~=cost-isint
1)证：(z+1)/(z-1)
=[(cost+1)+isint]/[(sint-1)+isint]
={2[cos(t/2)]^2+2isin(t/2)cos(t/2)}/{-2[sin(t/2)]^2+2isin(t/2)cos(t/2)}
=2cos(t/2)/[-2isin(t/2)]*[cos(t/2+sin(t/2)/[isin(t/2)+cos(t/2)]
=icot(t/2)所以(z+1)/(z-1)在|z|=1的前提下是纯虚数
2)|z^2-z+1|=|z(z-1+1/z)|=|z|*|z...全部
设z=cost+isint
--->|z|=1,1/z=z~=cost-isint
1)证：(z+1)/(z-1)
=[(cost+1)+isint]/[(sint-1)+isint]
={2[cos(t/2)]^2+2isin(t/2)cos(t/2)}/{-2[sin(t/2)]^2+2isin(t/2)cos(t/2)}
=2cos(t/2)/[-2isin(t/2)]*[cos(t/2+sin(t/2)/[isin(t/2)+cos(t/2)]
=icot(t/2)所以(z+1)/(z-1)在|z|=1的前提下是纯虚数
2)|z^2-z+1|=|z(z-1+1/z)|=|z|*|z+z~-1|
=1*|(cost+isint)+(cost-isint)-1|
=|2cost-1|
-1=-3=0=<|2cost-1|=<3
所以|z^2-z+1|的最大值是3，最小值是0。
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