How do i show that argument is continuous at points except its branch cut? Let $Arg(z)$ be the principal value of argument of $z$, so that $Arg:\mathbb{C}\setminus\{0\}\rightarrow (-\pi,\pi]$ is a function.
How do I prove that $Arg$ is continuous at $z$ for all nonnegative and nonzero $z$?
 A: For $z = x + iy$,
$$   
Arg(z) = \begin{cases} 
\arctan(\frac{y}{x}) &x>0  \\ 
\frac{\pi}{2}  & x = 0, \ y>0 \\ 
-\frac{\pi}{2}  & x = 0, \ y<0 \\
\pi +\arctan(\frac{y}{x})   & x < 0, \ y<0 \\
-\pi -\arctan(\frac{y}{x})   & x < 0, \ y>0 \end{cases}
$$
It's undefined for $z = 0$. 
We will prove continuity for $Arg(z)$, for the first case, $(x>0)$
In order to prove continuity at a point $z$ you must prove the following
$$
\lim_{\Delta z\to 0}Arg(z + \Delta z) - Arg(z) = 0  
$$
Where $\Delta z = \Delta x + i\Delta y$
Using the arctangent definition for $x>0$,
$$
Arg(z + \Delta z) - Arg(z) = \arctan\left(\frac{y + \Delta y}{x + \Delta x}\right) - \arctan\left(\frac{y}{x}\right)
$$
Now using the fact that $arctan$ is an odd function, 
$$
- \arctan\left(\frac{y}{x}\right) = \arctan\left(-\frac{y}{x}\right)
$$
And the arctangent additon formula, with a little bit of algebra you get
$$
\arctan\left(\frac{y + \Delta y}{x + \Delta x}\right) + \arctan\left(-\frac{y}{x}\right) = \arctan\left(\frac{x\Delta y  - y\Delta x}{x^2 + y^2 + x\Delta x + y\Delta y}\right)
$$
When $\Delta z \to 0$, both $\Delta x \to 0$ and $\Delta y \to 0$ so 
$$
\lim_{\Delta z\to 0}Arg(z + \Delta z) - Arg(z) = \lim_{\Delta z\to 0} \arctan\left(\frac{x\Delta y  - y\Delta x}{x^2 + y^2 + x\Delta x + y\Delta y}\right) = 0   
$$
