Application of Open Mapping Theorem This was stated without proof in the complex analysis text I am reading (Complex Made Simple by Ulrich, page 107). I'm sure it's easy, but I'm tired and need a little help.
Let $f$ be nonconstant and holomorphic in some region $V$ and assume $f'$ is nonconstant. Define
$$\Omega = \left\{ \frac{f(z)-f(w)}{z-w}: z,w\in V, z\neq w \right\}.$$
Why is $\Omega$ open? I don't see how to use the hypothesis that $f'$ is nonconstant. 
To be clearer: I tried something like the answers below. But Ulrich specifically mentions that the hypothesis that $f'$ is nonconstant is needed in addition to the hypothesis that $f$ is nonconstant. So I feel I'm missing some subtle point about the derivative. 
Edit: After reading the answers below, the key point is $f'$ nonconstant implies the difference quotient is nonconstant. 

Given $w\in V$, define
$$g_w(z):V\setminus\{w\}\to\mathbb C,\quad z\mapsto \frac{f(z)-f(w)}{z-w}.$$
Since $f$ is holomorphic on $V$ and since $f'$ is not constant, $g$ is a nonconstant holomorphic function on $V\setminus\{w\}$. Noting that $V\setminus\{w\}$ is open, by open mapping theorem, $g_w(V\setminus\{w\})$ is an open subset of $\mathbb C$.
It follows that 
$$\Omega=\bigcup_{w\in V}g_w(V\setminus\{w\})$$
is open.
