Define $D^*=\mathbb{C} \backslash \lbrace z \in \mathbb{C}:\operatorname{Re}(z) \leq 0, \operatorname{Im}(z)=0 \rbrace$. Prove that $f(z)=\arg(z)$ is not analytic on $D^*$.
My proof is as follows:
Let $\arg(z)= \theta$. Then the function becomes $f(r,\theta)=\theta$. By using CR-equation in polar form, we obtain $ru_r=0=v_{\theta}$ and $u_{\theta}=1 \neq 0=-rv_r$. Hence, $f$ is not differentiable, which implies not analytic.
Is my proof correct?
Your proof is fine, but here is one that does not require the polar form of the CR equations.
$\arg(z)$ is constant on each radial line. If it were analytic, it's derivative would need to be zero on each radial line. If the zeroes of an analytic function have a limit point, the function must be $0$. This means that $\arg(z)$ would be constant, which is not so.
This function is real-valued and non-constant, thus it is not analytic.
This is due to the fact that a real-valued analytic function is constant, which in turn, is an implication of Cauchy-Riemann equations.
That's a perfectly good method. If you can use the CR-equation in polar form, that seems like the quickest and most sensible way to solve the problem.
