# Is it possible to show that $\operatorname{Arg}z$ is not an analytic function by using the Cauchy–Riemann differential equations?

I would like to show that $\operatorname{Arg}z$ is not an analytic function by using the Cauchy–Riemann differential equations.

After a quick search I've found this:

From Wikipedia: Alternatively, the principal value can be calculated in a uniform way using the tangent half-angle formula, the function being defined over the complex plane but excluding the origin:

$\operatorname{Arg}(x + iy) = \begin{cases} 2 \arctan \left( \frac{y}{\sqrt{x^2+y^2}+x} \right) & \qquad x > 0 \text{ or } y \ne 0 \\ \pi & \qquad x < 0 \text{ and } y = 0 \\ \text{undefined} & \qquad x = 0 \text{ and } y = 0 \end{cases}$

Unfortunately, I was not able to make further progress. My question: Is it possible to show that $\operatorname{Arg}z$ is not an analytic function by using the Cauchy–Riemann differential equations?

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$\operatorname{Arg}:\mathbb{C}\to\mathbb{R}$ ... – Neal Mar 26 '12 at 23:20
What is your definition of $arg(z)$? does it take values in $S^1$ or in $\mathbb{R}$ or in $\mathbb{R}\subset \mathbb{C}$? The way I think of the Cauchy-Riemann equations, you need to be able to apply them to functions with values in $\mathbb{C}$. On the other hand, I think of $arg(z)$ as taking values in $S^1$. Once this definition problem has been settled, I would sidestep this formula nonsense and compute Cauchy-Riemann on the composition $arg( \operatorname{e}^{s+it} ) = t$ or $=t (mod 2\pi)$. – Sam Lisi Mar 26 '12 at 23:26

As Neal points out in the comments, there's no need to use an explicit formula for $\arg$ because there is a more general, and very simple, principle at work here. Any complex-analytic function $\mathbb{C}\to\mathbb{R}$ must be constant, but the argument function is nonconstant. The Cauchy-Riemann equations tell us, because the imaginary part $v:=0$ is constant, that $u_x$ and $u_y$ are also $0$ and hence $u$ (the real part) is constant. This sidesteps any formal or technical work you would otherwise need to do here.