I've been thinking how to prove that an analytic function $f$ in the domain $D$ is a constant if $f'(z)=0$ in every point in $D$, but I haven't figured it out yet.
What I was thinking is to use Cauchy-Riemann equations, but it didn't work well…
If this is not true, I would like to know the counterexample…
Here is what I tried:
Let $f(z)=u(x,y)+iv(x,y)$
$\color{blue}{(1)}\text{(Cauchy-Riemann})\begin{cases}
u_y=-v_x\\
u_x=v_y
\end{cases}$
$\color{blue}{(2)}\lim\limits_{\Delta z \to 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z }=0$ for all $z_0$ in $D$
I'm stuck here…
Best Answer
In complex analysis, a domain per definition is an open connected set $D\subset {\mathbb C}$. Given two arbitrary points $z_0$, $z_1$ in such a domain $D$ there is a path $$\gamma:\quad t\mapsto z(t)\in D\qquad(0\leq t\leq1)$$ with $z(0)=z_0$, $\>z(1)=z_1$. The auxiliary function $$\phi(t):=f\bigl(z(t)\bigr)\qquad(0\leq t\leq1)$$ of the real variable $t$ has derivative $\phi'(t)=f'\bigl(z(t)\bigr)\>z'(t)\equiv0$; whence $\phi$ is constant. It follows that $$f(z_1)=\phi(1)=\phi(0)=f(z_0)\ .$$ As $z_0$, $z_1\in D$ were arbitrary, the claim follows.