How do I prove that $f'(z)=0$ implies $f$ is constant? Let $V$ be an open connected subset of $\mathbb{C}$.
Let $f:V\rightarrow\mathbb{C}$ be a function whose derivative is $0$ on $V$.
How do I prove that $f$ is a constant on $V$?
I know that $V$ is path-connected, but I don't know whether this helps.
 A: Use the Cauchy-Riemann equations:
$$
f(x + iy) = u(x, y) + iv(x, y)
$$
Where $u$ and $v$ are real valued functions.  We have the constraints:
$$
\frac{\partial u}{\partial x} = \frac{\partial v }{\partial y} \\
\frac{\partial u}{\partial y} = -\frac{\partial v }{\partial x} 
$$
Giving:
$$
f'(x + iy) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}
$$
If $f' = 0$ then we have that $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} = 0$.  This then entails that $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial y} = 0$.  If $\frac{\partial u}{\partial x} = 0$ then $u = \alpha(y)$.  If you do all of them then you find that $u = \alpha(y) = \beta(x)$ and $v = \gamma(y) = \varphi(x)$.  The only way that $\alpha(y) = \beta(x)$ for all values of $x$ and $y$ is if $\alpha$ and $\beta$ have no dependence on either $x$ or $y$ and thus are constants.  The same follows for $v$.  Therefore $u = A$ and $v = B$ and $f(z) = A + Bi$.
A: Consider a point $z_0 \in V$. Now consider any other point $w \in V$. Since $V$ is connected, it is possible to find a curve starting at $z_0$ and ending at $w$. Let's call this curve $C_w$. We then have
$$f(w) - f(z_0) = \int_{C_w} f'(z) dz = \int_{C_w} 0 \cdot dz = 0$$
Hence, we have $f(w) = f(z_0)$. This is true for any $w \in V$. Hence, $f(w) = f(z_0)$ and hence $f(w)$ is a constant in $V$.
A: Hint: Choose a path $\gamma$ connecting two given points in $V$, and evaluate
$$\int_{\gamma} f'(z) dz$$ 
two different ways.
A: You can connect any point to a point $z_0$ by a polygonal line. Along each segment you can write $f(z) = f(z_i + t (z_f - z_i))$, and write this as two real functions of the real variable $t$, one for the real part and one for the imaginary part. Write the derivative in terms of them, you see both components are zero along the segment, so both components are constant and the function itself is constant on that segment, and thus over each segment of the path, and so over the full region.
