Show that $\text{arg}(f(z))$ is a constant $\Rightarrow$ $f(z)$ is constant in $D$. The full question is as follows
Let $f(z)$ be an analytic function in a region $D$ and $f(z) \neq 0$ in $D$. Show that $\text{arg}(f(z))$ is a constant $\Rightarrow$ $f(z)$ is constant in $D$.
My approach would be to use Cauchy Riemann in terms of polar coordiantes.
In polar coordinates the Cauchy-Riemann equations become  $$\dfrac{du}{dr}=\dfrac{1}{r}\dfrac{dv}{d\theta} ~~,~~ \dfrac{dv}{dr} = -\dfrac{1}{r}\dfrac{du}{d\theta}$$
The derivative in polar version at a point $z$ whose polar coordinates are $(r,\theta)$ is then $$f^{'}(z) = e^{-i\theta}(\dfrac{du}{dr}+i\dfrac{dv}{dr}) = \dfrac{1}{r}e^{-i\theta}(\dfrac{dv}{d\theta}-i\dfrac{du}{d\theta})$$
So how do i go on from here?
Since $arg(f(z))$ is equivalent to the $\theta$ in question, can i just say that $v_\theta = u_\theta = 0$?
Any help would be appreciated.
 A: You can go ahead with your Cauchy-Riemann approach, but I also know of a geometric argument which I like a lot. Suppose $f(z)$ is nonconstant on $D$, and let $a \in D$ such that $f'(a) \neq 0$. By the definition of derivative, we know that $f(a + h) = f(a) + hf'(a) + o(h)$. 
Choose $\alpha$ in $\Bbb{C}$ so that when $\alpha$ is considered as a vector in $\Bbb{R}^2$, $\alpha f'(a)$ is a unit vector perpendicular to $f(a)$. Let $h = r \alpha$, with $r$ real and positive. Note that $arg(f(a) + hf'(a)) \neq arg(f(a)).$ As $r$ goes to zero we see that $f(a + h)$ lies in a disk of radius $o(r)$ around $f(a) + h f'(a)$. For $r$ sufficiently small, the radius of this disc is smaller than the distance between $f(a) + h f'(a)$ and $f(a)$.
But $f(a)$ is the closest point to $f(a) + h f'(a)$ on the line $arg(z) = arg(f(a))$, since $h f'(a)$ is perpendicular to that line. It follows that when $h$ is sufficiently small, the disk and the line do not intersect. But $f(a + h)$ lies in the disk, and therefore cannot have the same argument as $f(a)$. This is a contradiction.
We conclude that if $arg(f(z))$ is constant, $f'(z)$ is identically zero on $D$. Assuming that $D$ is connected this implies that $f$ is a constant function there.
A: I think you answer is correct. Write
$$f^{'}(z) =\dfrac{1}{r}e^{-i\theta}(\dfrac{dv}{d\theta}-i\dfrac{du}{d\theta})=\dfrac{1}{r}e^{-i\theta}\dfrac{-idf}{d\theta}=\dfrac{-i}{r}e^{-i\theta}\dfrac{df}{d\theta}$$
from the $argf$ is constant it conclude $f'(z)=0$ on $D$. Thus $f$ is constant on $D$.
A: To give another proof:
By the open mapping theorem, $f(D)$ must be open, if $f$ is not constant. If $\arg(f(z))$ is constant, then $f(D)$ cannot be open.
A: Let $\arg f(z)=\theta_0$, and then
$$f(z)=|f(z)|e^{\rm{i}\theta_0},\qquad \forall z\in D.$$
Let
$$F(z)=e^{-\rm{i}\theta_0}f(z)=|f(z)|\in \mathbb R,\qquad \forall z\in D,$$
then $F$ is real-valued holomorphic function $(\text{Im} F(z)=0)$,
which implies $f$ is constant.
