 Odd square-root transform of a normalized univalent function

Because $f(z)/z$ is holomorphic and nowhere zero on the unit disc, its square root with value one at zero exists. Defining $h(z)=z\sqrt{f(z^2)/z^2}$ gives an odd normalized univalent function. If $h(z)=h(w)$, squaring gives $z^2=w^2$; the possibility $z=-w$ and oddness force both to be zero. Applying the area theorem for univalent functions to $1/h(1/\zeta )$ yields $|a_2|\le 2$.