that $\lim_{n\to\infty}\frac{\Im f(z_n')}{
\Im z_n'}=c$. Clearly $\Im f(z_n')\to0$ as $n\to\infty$, and $\{\Re
f(z_n')\}_{n\in\mathbb N}$ is a bounded sequence in $\mathbb R$. Moreover, if $\{z_n\}_{n\in\mathbb N}$ and $\{z_n'\}_{n\in\mathbb N}$ are two arbitrary sequences converging to $\alpha$ along which $\Im f(z)/\Im z$ stays bounded, then $\{\Re(f(z_n)-f(z_n'))\}_{
n\in\mathbb N}$ converges to zero. This implies that $\lim_{n\to\infty}
f(z_n)$ exists for any sequence $\{z_n\}_{n\in\mathbb N}$ such that $\{\Im f(z_n)/\Im z_n\}_{n\in\mathbb N}$ is bounded and $\lim_{n\to
\infty}z_n=\alpha$. We agree to call this limit $f(\alpha)$. Taking limit along $z_n'$ in we obtain $$\left|\frac{f(z)-f(\alpha)}{z-\alpha}\right|^2\leq
c\frac{\Im f(z)}{\Im z},\quad z\in\mathbb C^+.$$ Fix an $M\in[0,+\infty)$. Let $D_M=\{z\in\mathbb C^+\colon
|\Re z-\alpha|\leq M\Im z\}$. For any $z\in D_M$, this implies $$\begin{aligned}
\left(\Re f(z)-f(\alpha)\right)^2 & \leq & c\frac{\Im f(z)}{\Im z}|z-
\alpha|^2-\left(\Im f(z)\right)^2\\
& = & \Im f(z)\left(\frac{c|z-\alpha|^2}{\Im z}-\Im f(z)\right)\\
& = & \Im f(z)\left(c\Im z\frac{|\Re z-\alpha|^2}{(\Im z)^2}
+c\Im z-\Im f(z)\right)\\
& \leq & \Im f(z)\left(c(M^2+1)\Im z-\Im f(z)\right).\end{aligned}$$ We conclude that $\Im f(z)/\Im z\leq c(M^2+1)$ for all $z\in D_M$ and thus $\displaystyle\lim_{\stackrel{z\longrightarrow0}{{\sphericalangle}}}
f(z)=f(\alpha)$. Moreover, for $M=0$ (i.e. $z$ of the form $\alpha+iy$) we have $c\geq\Im f(\alpha+iy)/y$, which together with the definition of $c$ implies $\lim_{y\downarrow0}\frac{\Im f
(\alpha+iy)}{y}=c$, so that $$\frac{\left(\Re f(\alpha+iy)-f(\alpha)\right)^2}{y^2}\leq \frac{\Im f(\alpha+iy)}{y}
\left(c-\frac{\Im f(\alpha+iy)}{y}\right)\to0\quad\text{as }y\downarrow0.$$ These two facts imply, via direct computation, that $\lim_{y\downarrow0}\frac{f(\alpha+iy)-f(\alpha)}{iy}=c$. 