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.

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 $D$.
