原来u,v是f(z)的实部和虚部(这个在题目处最好说明一下,不然会误解).
f(z)=u+iv=(2xy-2y)+i(2x-x^2+y^2+c)
=2(x-1)y+i(y^2-(x-1)^2+c+1)
=i{y^2-2i(x-1)y+[i(x-1)]^2+c+1}
=i{[y-i(x-1)]^2+c+1}
=i{[-i(iy+x-1)]^2+c+1}
=i{(-i)^2(z-1)^2+c+1}
=-i(z-1)^2+i(c+1) 原来u,v是f(z)的实部和虚部(这个在题目处最好说明一下,不然会误解).
f(z)=u+iv=(2xy-2y)+i(2x-x^2+y^2+c)
=2(x-1)y+i(y^2-(x-1)^2+c+1)
=i{y^2-2i(x-1)y+[i(x-1)]^2+c+1}
=i{[y-i(x-1)]^2+c+1}
=i{[-i(iy+x-1)]^2+c+1}
=i{(-i)^2(z-1)^2+c+1}
=-i(z-1)^2+i(c+1)