Does this set of functions separate points of $\Omega$? Let $\Omega$$\subset\mathbb{C}$ be a open connected set (domain). Let $z\in\Omega$ I want to show that if all bounded holomorphic functions on $\Omega$ are not just the constant functions (denoted by $H^\infty(\Omega)\not\equiv\mathbb{C}$), then, for every $0\neq\xi\in\mathbb{C}$, there exists a bounded holomorphic function $f$ on $\Omega$ such that $f’(z)\xi\neq 0$
The proof goes like this (see proof of Proposition 2.5.1 in this book).
Suppose $H^\infty(\Omega)\not\equiv\mathbb{C}$ and let $f_0\in H^\infty(\Omega), a_1,a_2\in\Omega$, such that $f_0(a_1)\neq f_0(a_2)$.
Define for $j=1,2$, $f_j$ as 
\begin{equation*}
    f_j(z) = \begin{cases}
              \frac{f_0(z)-f_0(a_j)}{z-a_j}& \text{if } z\neq a_j,\\
               f_0’(a_j) & \text{if } z= a_j.
          \end{cases}
\end{equation*}
Then it is said that $f_0,f_1,f_2$ separates points of $\Omega$. And $rank(f_0’,f_1’,f_2’)=1$ on $\Omega$ and hence proved.
I do not understand how $f_0,f_1,f_2$ separates points of $\Omega$? And what is the meaning of $rank(f_0’,f_1’,f_2’)=1$?
 A: The statement about separating points means that for any distinct $z,w \in \Omega$ there exist $j \in \{0,1,2\}$ with $f_j(z) \ne f_j(w)$, and the statement about the rank of the derivatives just means that these three derivatives do not simultaneously vanish at any point in $\Omega$. And since you are working in one complex dimension here, you don't need $\xi$, just need to show that given $z_0 \in \Omega$ (which you omitted from your assumptions, but this is the way it is phrased in the referenced book) there exists a bounded holomorphic function $f$ in $\Omega$ with $f'(z_0) \ne 0$.
I think there is a more elementary proof than the one in the book. By assumption there exists a non-constant function $f$ in $\Omega$. If $f'(z_0) \ne 0$, then you are done. Otherwise, the Taylor expansion at $z_0$ looks like $f(z) = w_0 + a_n (z-z_0)^n + \ldots$ with $w_0 = f(z_0)$ and $a_n \ne 0$ for some $n \ge 2$. Then $g(z) = \frac{f(z)-w_0}{(z-z_0)^{n-1}}$ is another bounded function in $\Omega$ (which is not hard to show) with $g(z) = a_n (z-z_0) + \ldots$, so $g'(z_0) = a_n \ne 0$.
