Find the image of the following function:
I solved this but I am not 100% sure.
if w=z/(1-z)^2 then for z!=1 then wz^2-(2w+1)z + w=0 and therefore
z=(2w+1 + sqrt(4w+1))/2w or z=(2w+1 - sqrt(4w+1))/2w for every w!=0
and for w=0 we can pick z=0.
So for every w in C there exists a z in C so w=f(z) and therfore the image of f(z) is the entire complex plane C.
I can't seem to contradict this, but still it dosent seem right to me...
Would appreciate any comments