Plessner's theorem and radial limit of derivative of univalent function I'm reading an article by Pommerenke (Conformal mapping and linear measure) and I have one question about it. First the assumptions:
Let $f$ be analytic and univalent in the unit disk $D$ and let $A_1 = \{\xi : |\xi|=1,\, f'(\xi)\mbox{ exists, is }\neq 0, \infty\}$,
where $f'(\xi)$ denotes the angular derivative (nontangential limit of $f'$).
By Plessner's theorem, for almost all $\xi$ in the boundary of $D$ there either exists $f'(\xi) \neq \infty$ or else $f'(z)$ has every point in $C$ as a limit point as $z \to \xi$ in Stolz angle terminating at $\xi$.
Now by the definition of $A_1$, for almost every $\xi \in A_1^c$ there exists a sequence $z_n$ such that
$z_n \to \xi, \quad f'(z_n)\to 0 \quad$ as $n\to\infty.$
Now it says in the article that "Since $f$ is univalent we conclude that $f'(|z_n|\xi)\to 0$ as $n\to\infty$." Why does this hold? How the univalentnes imply that the limit along some sequence is same as radial limit when approaching the point $\xi$ with same speed? 
 A: Definition. A holomorphic function $g:D\to\mathbb C$  is a Bloch function if $$\|g\|_{\mathcal B}:=\sup_{z}(1-|z|^2)|g'(z)|<\infty$$ 
Suppose $f$ is univalent in $D$. The function $\log f'$ has a single-valued branch in $D$ because $f'$ does not vanish. It does not matter what branch we use below, because they differ by a constant. 
Claim 1. If $f$ is univalent, then $\|\log f'\|_{\mathcal B}\le 6$.
This is well-known but also easy to prove. Let $g=\log f'$. Then $g'=f''/f'$. For a fixed $a\in D$ consider the function 
$$\frac{f\left(\frac{z+a}{1+\bar a z}\right)-f(a)}{(1-|a|^2)f'(a)} = z+c_2z^2+\dots \quad \text{where} \quad c_2=\frac12(1-|a|^2)\frac{f''(a)}{f'(a)}-\bar a$$
Since this function is univalent in $D$, we have $|c_2|\le 2$. Thus, 
$$\left|(1-|a|^2)\frac{f''(a)}{f'(a)}-2\bar a\right|\le 4$$ which implies 
$\left|(1-|a|^2)\frac{f''(a)}{f'(a)}\right|\le 6$. $\Box$
Claim 2. If $g$ is a Bloch function, then for every $z\in D$ and  $\theta\in\mathbb R$ we have $$|g(e^{i\theta}z)-g(z)|\le \frac{|\theta|}{1-|z|^2}\|g\|_{\mathcal B} \tag1$$
Proof: integrate $g'$ along the circular arc connecting $z$ to $e^{i\theta}z$. The length of this arc is less than $|\theta|$. Estimate $|g'|$ by $\|g\|_{\mathcal B}/(1-|z|^2)$. $\Box$
In our situation $z=z_n$ and $e^{i\theta }z = |z_n|\xi$ both lie within the same Stolz angle with vertex at $\xi$. Therefore, $|\theta|\le M(1-|z|^2)$ where $M$ is a constant depending on the size of the Stolz angle. By (1), $$|\log f'(z_n)-\log f'(|z_n|\xi)|\le 6M$$ for all $n$. Since $\operatorname{Re}\log f'(z_n)\to-\infty$ , it follows that $\operatorname{Re}\log f'(|z_n|\xi)\to-\infty$. $\Box$

tl;dr version: "It is easy to see that" 

