Question

已知w=z+i(z∈C),(z-2)/(z+2)是纯虚数...

Answer 1

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