Why does $w(z)=z^2$ represent a conformal map from the $z$ plane to the $w$ plane?

Question

Doesn't every straight line $y=kx$ in the $z$-plane become a straight line $u=v\cdot2m/(1-m^2)$ in the $w$-plane such that each point having a phase angle $\theta$ in the $z$-plane has a phase angle $2\theta$ in the $w$-plane? This would mean that if the angle between two straight lines in the $z$-plane is $\alpha-\beta$, the corresponding angle between the corresponding lines in the $w$-plane should be $2\alpha-2\beta=2(\alpha-\beta)$. Then, how come is the angle preserved?

Answer 1

The angle is not preserved. And that's not a problem since $w'(0)=0$. The angles must be preserved at those points $z$ at which $w'(z)\ne0$.