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For $(x,y)\in\Bbb R^2$ with $(x,y)\neq(0,0)$, let $\theta=\theta(x,y)$ be the unique real number such that -$\pi\lt\theta\le\pi$ and $(x,y)=(r\cos\theta, r\sin\theta)$, where $r=\sqrt{x^2+y^2}$. Then is the resulting function $\theta:\Bbb R^2-\{(0,0)\}\to \Bbb R$ continuous and differentiable?

I have confusion about the definition of the function theta. How to check the continuity and differentiability of this function.

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    $\begingroup$ $\theta(x,y)$ is the angle that the line from $(0,0)$ to $(x,y)$ makes with the positive $x$-axis. $\endgroup$ – Arthur May 17 '17 at 15:40
  • $\begingroup$ yes, now it is easy to check the continuity and differentiability of this function. thank you sir you have cleared my confusions $\endgroup$ – Priyanka May 17 '17 at 16:42
  • $\begingroup$ What happens at $(-1,0)$? Is it continuous there? $\endgroup$ – Arthur May 17 '17 at 16:43
  • $\begingroup$ Sir i don't know whether it is continuous at (-1,0) or not. If you know then please let me know the answer $\endgroup$ – Priyanka May 18 '17 at 12:35
  • $\begingroup$ Just telling you the answer has very little value. I'm of the belief that if you find the answer yourself, through a little guidance, you will come away from it wiser than if I had told you the answer. So, without further ado: for a very small, positive number that I shall call $\epsilon$, what is $\theta(-1, \epsilon)$? What is $\theta(-1, -\epsilon)$? $\endgroup$ – Arthur May 18 '17 at 13:02

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