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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 $z$?

Rubertos
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  • Hint: $\arctan$ is continuous. You'll need to do some patching. – Ted Shifrin Sep 13 '14 at 11:34
  • @Ted would you please give me some more hints? I have tried this more than an hour, but still haven't proved anything.. It's geometrically trivial, but I don't know how to actually prove it.. – Rubertos Sep 13 '14 at 11:39

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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 $$ \lim_{\Delta z\to 0}Arg(z + \Delta z) - Arg(z) = \lim_{\Delta z\to 0} \arctan\left(\frac{x\Delta y - y\Delta x}{x^2 + y^2 + x\Delta x + y\Delta y}\right) = 0 $$

vladimirm
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