The conformal map from interior of ellipse to interior of the unit disk (property check)

Question

Based on example 5 (page 546) and example 7 (page 550) of the book "Applied and computational complex analysis. Volume 3, Wiley, 1986" written by Peter Henrici, if $a,b>0$ satisfies $a^2-b^2 = 1$ and we define $\rho>1$ through $\rho+\rho^{-1} = 2a$ (or through $\rho - \rho^{-1} = 2b$), the conformal map sending the interior of the elliptical region $R:=\{(x,y)\in \mathbb{R}^2 \mid \frac{x^2}{a^2}+\frac{y^2}{b^2} \leq 1\}$, which in polar coordinates reads $$R = \left\{(\theta,r) \mid r \leq \frac{ab}{\sqrt{a^2\sin^2\theta + b^2\cos^2\theta}}:=r(\theta)\right\},$$ to the interior of the unit circle is given by \begin{equation}\label{E2D} h(z) = \frac{4}{h'(0)}\sum_{m=0}^\infty \frac{(-1)^mT_{2m+1}(z)}{\rho^{4m+2}-\rho^{-4m-2}},\quad z\in R \setminus \partial R, \end{equation} where $(T_n)_{n\geq 0}$ are the Chebyshev polynomials of the first kind and $$h'(0) = \left(4\sum_{m=0}^\infty \frac{2m+1}{\rho^{4m+2}-\rho^{-4m-2}}\right)^{\frac 12}.$$ But I get stuck when trying to check this is indeed the correct conformal map. Specifically, I do not know how to show $$|h(z)| < 1 \quad \text{for $z = re^{i\theta} \in R \setminus \partial R$}.$$ Also, another key question that is of my interest concerns the boundary behavior of $h$: what is the image under $h$ of the boundary "elliptical" arc $\{(\theta,r(\theta)) \in E \mid |\theta| \leq \delta \}$ (where $E := \partial R$ is boundary of the elliptical region enclosed by $R$ and $\delta$ is small)? Thanks for any help or suggestions!