Univalence of regular branches

If $f(z)$ is a univalent (1-1 analytic) function with domain the unit disc such that $f(z) = z\sum_{n=2}^{\infty } a_n z^n$, then prove that

is a univalent function in the unit disc.

Let the function $w = f(z)$ effect a schlicht mapping of the unit circle $\left|z\right| < 1$ onto a region (non-empty connected open set) $\mathfrak {R}$ and let $\varphi (w)$ be schlicht in $\mathfrak {R}$. Then $\varphi [f(z)]$ is schlicht in $\left|z\right| < 1$.

Let the function $f(z)$ be schlicht in the unit circle $\left|z\right| < 1$ and let $f(0) = 0$. Then the function

with a fixed branch of the square root, is also schlicht in $\left|z\right| < 1$. A similar result holds for $\sqrt[n]{f(z^{n})}$, where $n$ is a positive integer.

The function $\varphi (z)$ in problem 79 is the most general odd function that induces a schlicht mapping of the unit circle $\left|z\right| < 1$. More precisely: If $\varphi (z)$ is an odd function schlicht for $\left|z\right| < 1$, then there exists a function $f(z)$ which is schlicht for $\left|z\right| < 1$ from which $\varphi (z)$ is derived as in problem 79.

If $f \in \mathscr {S}$ (Definition 14.10) and $n$ is a positive integer, prove that there exists a $g \in \mathscr {S}$ such that $g^{n}(z) = f\left(z^{n}\right)$ for all $z \in U$.

Prove that if the function $\operatorname {Ln} z$ admits the isolation of a regular branch in the domain $D$, then this regular branch defines a conformal mapping of the domain $D$.

Let $n$ be a positive integer. Prove that if the function $\sqrt[n]{z}$ admits the isolation of a regular branch in the domain $D$, then this regular branch defines a conformal mapping of the domain $D$.

Let the function $f(z)$ be regular and define a conformal mapping of the disk $\left\{ z:\left|z\right| < 1\right\}$, with $f(0) = 0$. Prove that the multivalued mapping $\left\{ \sqrt[n]{f\left(z^{n}\right)}\right\}$ in the disk $\left\{ z:\left|z\right| < 1\right\}$ splits into $n$ functions that are regular and define conformal mappings in the disk.

Let $-\infty < a_{1} < \ldots < a_{n} < +\infty$. Prove that any regular branch of the multivalued function $f(z) = \sqrt[n]{\left(z-a_{1}\right) \ldots \left(z-a_{n}\right)}$ in the half-plane $\left\{ z: \mathrm{Im}\left(z\right) > 0\right\}$ defines a mapping that is conformal in this half-plane.