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On what curve is $\arg (z)$ discontinuous if it is defined as the value of $Arg(z)$ satisfing the inequality: $|z|-2\pi<\operatorname{Arg}(z)\leq|z|$

would it be a ray from the origin with argument equal to $|z|$? Could somebody clarify this for me?

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    The Wikipedia article on the [argument](http://en.wikipedia.org/wiki/Argument_%28complex_analysis%29) has the convention the other way around: $\operatorname{Arg}$ is a canonical value, and $\arg$ is the set of all values. (I followed your convention in my answer.)2012-09-28

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It's discontinuous on the curve $\operatorname{Arg}(z)=|z|$. This is an Archimedean spiral.