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I know how to show the function $ArgZ$ is not continuous on negative real axis. But I can not prove how it is continuous else where. Can anyone enlighten me?

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    See https://math.stackexchange.com/questions/271117/continuity-of-arg-z and https://math.stackexchange.com/a/624840/108128. – Nosrati Aug 13 '17 at 02:46

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Note that $$ \begin{align} \frac{\sin(\theta)}{1+\cos(\theta)} &=\frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)}\\[12pt] &=\tan(\theta/2)\tag{1} \end{align} $$ Using $(1)$, $\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}$, and $\cos(\theta)=\frac{x}{\sqrt{x^2+y^2}}$, we get that $$ \begin{align} \arg(x+iy) &=\theta\\ &=2\arctan\left(\frac{y}{\sqrt{x^2+y^2}+x}\right)\tag{2} \end{align} $$ which is continuous for all $\mathbb{C}$ away from the negative real axis (the negative real axis is precisely where $\sqrt{x^2+y^2}+x=0$).

robjohn
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  • This is parallel to the this answer. The only difference being the branch of $\arg$ used. Here it is the branch from $-\pi$ to $\pi$ and there it is the branch from $0$ to $2\pi$. – robjohn Aug 13 '17 at 15:15