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How do i show that argument is continuous at points except its branch cut?

I posted a question to ask whether the principal value of Argument $Arg:\mathbb{C}\setminus \{0\}\rightarrow (-\pi,\pi]$ is continuous at every nonnegative nonzero number.

Then, vladimirm answered my question so now I'm checking his argument.

However, I'm not sure that whether the function $Arg$ is differentiable at every nonnegative nonzero number.

Is it differentiable? Then how do i prove it?

Community
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Rubertos
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1 Answers1

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The formula $$\tan{\theta\over2}={\sin\theta\over 1+\cos\theta}={y\over\sqrt{x^2+y^2}+x}\qquad(-\pi<\theta<\pi)$$ allows to write the principal value of the argument in the form $${\rm Arg}(x,y)=2\arctan{y\over\sqrt{x^2+y^2}+x}$$ (note that the denominator is nonzero in the considered domain $\Omega$). Since the right hand side has continuous partial derivatives with respect to $x$ and $y$ on $\Omega$ it follows that ${\rm Arg}$ is indeed differentiable on $\Omega$.

Its worthwhile to memorize the following formula: $$\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2}, {x\over x^2+y^2}\right)\ ,$$ which is valid in the full punctured plane.

Christian Blatter
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  • So $Arg$ is not differentiable? It's now clear that it is continuous but I dunno why it's trivial to determine its differentiability – Rubertos Sep 13 '14 at 15:00
  • what is $\Omega$ here? It is, perhaps, the punctured plane for all $(x,y)\in\Bbb R^2$ such that $\theta\neq\pi$? – Masacroso Dec 21 '17 at 01:04
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    @Masacroso: $\Omega$ is the complex plane with the negative real axis together with the origin removed. – Christian Blatter Dec 21 '17 at 08:04