Since $$\left|\frac{f(z)-f(\alpha)}{z-\alpha}\right|^2\leq
c\frac{\Im f(z)}{\Im z}\leq c^2(M^2+1),\quad z\in D_M, M\ge0,$$ it follows straightforwardly that $$\lim_{\stackrel{z\longrightarrow0}{{\sphericalangle}}}
\frac{f(z)-f(\alpha)}{z-\alpha}=c$$ (see for example [@garnett Exercise 5, Chapter I]).

Considering the classical definition of the derivative, the above directly implies that $\lim_{y\downarrow0}f'(\alpha+iy)=c.$ Relation implies that $|f'(z)|\leq c(M^2+1)$ for $z\in
D_M$, so, by the same [@garnett Exercise 5, Chapter I], $\displaystyle\lim_{\stackrel{z\longrightarrow0}{{\sphericalangle}}}
f'(z)=c$. This proves (1).

To prove (2), simply observe that $$\left|\frac{f(\alpha+iy)-f(\alpha)}{iy}\right|^2=
\left|\frac{(\Re f(\alpha+iy)-f(\alpha))^2+(\Im f(\alpha+iy))^2}{y^2}
\right|\ge\frac{(\Im f(\alpha+iy))^2}{y^2},$$ so that $\liminf_{z\to\alpha}\frac{\Im f(z)}{\Im z}<\infty$. Part (2) follows now from part (1).

To prove part (3), we apply the classical mean value theorem to bound $\Im f(\alpha+iy)/y$. The result follows then from part (1).

We feel it necessary to reiterate that no claim to novelty is made for this proof, and we chose to write it down here for the sake of making the paper more self-contained.

Proof of the main result
========================

In this section we prove Theorem \[Main\]. The proof makes use quite often of the results, and sometimes of the proof, of Theorem \[JC\]. For the sake of simplicity, we will isolate some elements of the proof in separate lemmas.

For any $n\in\mathbb N$ and any state $\varphi$ on $M_n(\mathcal A)$, $z\mapsto\varphi(f(\alpha+zv))$ is a self-map of $\mathbb 