求助一道高中数学题·高手来下谢谢

设z∈C，且|z|=1，当|（z-1）（z-i）|最大时，z=
解：∵│z│=1,∴可设z=cosα+isinα
故 z-1=cosα-1+isinα; z-i=cosα+i(sinα-1)
│(z-1)(z-i)│=│z-1││z-i│={√[(cosα-1)²+sin²α]}{√[cos²α+(sinα-1)²]}
=√{[2(1-cosα)][2(1-sinα)]}=2√[(1-cosα)(1-sinα)]=2√{[2sin²(α/2)][1-cos(π/2-α)]}
=2√{[2sin²(α/2)][2sin²(π/4-α/2)]}=4│sin(α/2)sin(π/4-α/2)│
=4│-(1/2)[cos(π/4)-cos(α-π/4)]│=2│√2/2-cos(α-π/4)│≤2+√2
当α-π/4=π,即α=5π/4时等号成立。此时
z=cos(5π/4)+isin(5π/4)=-cos(π/4)-isin(π/4)=-[(√2)/2](1+i)]

原式＝|1-(1+i)z+i|=|(1-z)(1+i)|,运用向量，＝|(1-sinα,-cosα)(1,1)|=|sinα+cosα-1|=|√2sin(α+π/4)-1|,当取2kπ+5π/4时最大，所以z=sinα+cosα=-√2/2-√2/2i.over

设z=x+yi
|z-1|=1
(x-1)2+y2=1
设x-1=cosa ,y=sina

|z-i|
=√[x2+(y-1)2]
=√[(cosa+1)2+(sina-1)2]
=√(cos2a+2cosa+1+sin2a-2sina+1)
=√[3+2(cosa-sina)]
=√[3+2√2(cos(a+π/4)]
所以最大值=√（3+2√2)=√2+1

令Z=ai+b然后代入|（z-1）（z-i）|
注意a、b的限制条件 |z|=1