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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 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.) – joriki Sep 28 '12 at 11:56
up vote 1 down vote accepted

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

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Thankyou @joriki – Mykolas Sep 28 '12 at 12:07

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