 Solution

For an injective holomorphic function, the real Jacobian determinant is $|h^{\prime }{|}^2$. The area formula for a univalent power series, obtained by change of variables, is thereforeIndeed, insert $h^{\prime }(\rho e^{i\theta })={\sum }_{n\ge 1}n{c}_n{\rho }^{n-1}{e}^{i(n-1)\theta }$. Orthogonality eliminates the cross terms in the angular integral, and radial integration gives $2\pi n^2{\int }_0^r{\rho }^{2n-1}d\rho =\pi n{r}^{2n}$. The power series converges uniformly on each smaller closed disk, justifying the calculation for every $r<1$.

Since $|h(z){|}^2=|\varphi (z^2)|$, the Koebe growth theorem gives $|h(z)|\le |z|/(1-|z{|}^2)$. Hence $h(r\mathbb{D})$ lies in the disk of radius $r/(1-r^2)$, so