To prove that $f(z) = constant$ if $f'(z) = 0$, why is it necessary to prove that $u$ and $v$ are constant for all paths?

Question

In Churchill's "Complex Variables and Applications", when proving the following statement:

$f'(z) = 0 \space\space \forall z \in \mathbb{D}\subset \mathbb{C} \implies f(z) = constant \space\space \forall z \in \mathbb{D}$

First we write $f = u + iv$. Since $f'(z)$ exists in all points on the domain and it equals $0$, then $u'_{x} + i v'_{x} = 0$ and, because it fulfills the Cauchy-Riemann equations, $v'_{y} - i u'_{y} = 0$.

So $u'_{x} = u'_{y} = v'_{x} = v'_{y} = 0$.

Then he goes on proving that for every path between any two points in $\mathbb{D}$, the directional derivatives of $u$ and $v$ are $0$ and thus $u(x,y) = a$, $\space\space$ $v(x,y) = b$ and $f(z) = a + ib = constant$ for all $z \in \mathbb{D}$.

I don't understand why this last step is necessary. If the derivatives of $u$ and $v$ are already known to be $0$, doesn't that already imply that $u$ and $v$ must be constant, and so is $f$?

I apologize if this is too basic, but I'm very rusty on my multivariable real calculus.

Answer 1

I agree with you. If $u_x$ and $u_y$ are both the null function, then $u$ is constant. The same argument applies to $v$. Therefore, $f$ is constant.

Answer 2

The problem here is that the C-R equations are only concerned with derivatives in the $x$ and $y$ directions but none of the infinitely many directions inbetween. Therefore it doesn‘t follow entirely trivially that $f$ is constant.

However, we can assume the set that $f$ is defined on to be open and locally path connected. It is easy to show that any two points $z_0$ and $z_1$ in a path component can be joined by a sequence of horizontal and vertical paths. Along each of those sub-paths the change in $f$ is zero, so going along the path we get $f(z_0) = f(z_1)$. Thus $f$ is constant on the path component, i.e. locally constant.