Prove that $f$ is constant if $f'=0$ Suppose that $f$ is holomorphic on a domain $D$ and $f'=0$ on $D$. Prove that $f$ is constant on $D$.
 A: First write $f=u+iv$ where $u,v$ are functions from $\Bbb{R}^2\to\Bbb{R}$. Since $f'=0$ and by the Cauchy-Riemann equations, we have $u_x+iv_x=0=v_y-iu_y$. Then $u_x=u_y=v_x=v_y=0$. I will prove that $u$ remains constant the proof that $v$ remains constant is analogous.
Take points $z,z' \in D$ on a segment in $D$ and let $s$ be the distance from $z $to $z'$ along this segment and let $w$ be a unit vector on this segment pointing at the direction of increasing $s$, then the directional derivative $\dfrac{du}{ds}=\nabla u \cdot w$ however $\nabla u =0$ then $\dfrac{du}{ds}=0$ thus $u$ is constant on the segment. Now any $p\in D$ can be joined to $z$ by finitely many polygonal lines, on which as we have seen $u$ is constant, then $u(z)=u(p)$.
A: And yet another of doing so: Precomposing a non-degenerate line segment $γ\colon [0..1] → D$, we have $(f∘γ)' = 0$ by the chain rule, so $f∘γ$ is constant by the fundamental theorem of calculus. Therefore, $f$ is constant on a line segment and by the identity theorem constant on all of $D$.
A: Take $c\in D$ and write f as its Taylor series centrated in $c$. Let $r=dist(c,\partial D)$. Hence you know that $f(z)=\sum_{n=0}^{+\infty}\frac{f^{(n)}(c)}{n!}(z-c)^n\;\;\forall z\in B_{\mathbb C}(c,r[$. But $f'\equiv 0$ on $D$ by hypotesis, hence $f^{(n)}(c)=0\;\;\forall n\geq1$. Hence $f\equiv f(c)$ on $B_{\mathbb C}(c,r[$. But for the arbitrarity of $c\in D$ you desume that $f$ is constant on the whole $D$ (that is connected by definition, since it's a domain).
