已知w=z+i(z∈C),(z-2)/(z+2)是纯虚数...已知w=z+i(z∈C),且z-2/z+2为纯虚数,又│w+1│^2+│w-1│^2=16,求w的值.

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已知w=z+i(z∈C),(z-2)/(z+2)是纯虚数...已知w=z+i(z∈C),且z-2/z+2为纯虚数,又│w+1│^2+│w-1│^2=16,求w的值.已知w=z+i(z∈C),(z-2)

已知w=z+i(z∈C),(z-2)/(z+2)是纯虚数...已知w=z+i(z∈C),且z-2/z+2为纯虚数,又│w+1│^2+│w-1│^2=16,求w的值.
已知w=z+i(z∈C),(z-2)/(z+2)是纯虚数...
已知w=z+i(z∈C),且z-2/z+2为纯虚数,又│w+1│^2+│w-1│^2=16,求w的值.

已知w=z+i(z∈C),(z-2)/(z+2)是纯虚数...已知w=z+i(z∈C),且z-2/z+2为纯虚数,又│w+1│^2+│w-1│^2=16,求w的值.
设z=a+bi;
w=z+i=a+(b+1)i;
z-2=(a-2)+bi;
z+2=(a+2)+bi;
(z-2)/(z+2)=[(a-2)(a+2)+b^2]/2b^2+[b(a+2)-(a-2)b]/2b^2i;
(z-2)/(z+2)是纯虚数;
所以:
[(a-2)(a+2)+b^2]/2b^2=0;a^2+b^2=4;---(1)

b(a+2)-(a-2)b≠0;b≠0;
│w+1│^2=|(a+1)+(b+1)i|^2=(a+1)^2+(b+1)^2;--(2)
│w-1│^2=|(a-1)+(b+1)i|^2=(a-1)^2+(b+1)^2;----(3)
(2)+(3):
2(a^2+b^2)+4b+4=16;
代入(1)
2*4+4b+4=16;b=1;
将a=1带入(1);
a=±√3;
z=±√3+i;
w=±√3+2i