Let $r\in\mathbb{R}^{+}$, $\theta\in\mathbb{R}$ and $z_{0}\in\mathbb{C}$.
Does $\arg{(r\text{e}^{i\theta}+z_{0})}\longrightarrow\theta$ as $r\longrightarrow\infty$?
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2Maybe try using the fact that if $z\in \mathbb{C}$ and $\lambda\in \mathbb{R}^+$ then $\arg(\lambda z) = \arg(z)$. Thus $\arg(re^{i\theta} + z_0) = \arg(e^{i\theta} + z_0/r)$, right? – froggie Jul 28 '15 at 12:17
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Right, but does $\lim$ and $\arg$ commute? – BasicUser Jul 28 '15 at 12:20
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It depends what you mean by $\arg$. You cannot define $\arg$ in a continuous single-valued manner on $\mathbb{C}^$. If you are thinking of $\arg$ as a single-valued function, then it must have discontinuities, which means exactly that there will be points where the $\lim$ and $\arg$ will not commute (this is Ron Gordon's answer). However you can define $\arg\colon \mathbb{C}^\to \mathbb{R}/2\pi\mathbb{Z}$ continuously, and by continuity $\lim$ and $\arg$ will commute. – froggie Jul 28 '15 at 12:30
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Not most of the time. Let $z_0=x_0+i y_0$. Then
$$\arg{(r e^{i \theta}+z_0)} = \arctan{\left [ \frac{r \sin{\theta}+y_0}{r \cos{\theta}+x_0} \right ]} $$
As $z_0$ is fixed, then as $r \to \infty$, the ratio becomes $\tan{\theta}$. So if $\theta \in [-\pi/2,\pi/2]$, the answer is yes. However, failing that, one may extend the domain of the arctangent to $\theta \in [-\pi,\pi]$ by tracking the quadrant in which $e^{i \theta}$ lies. Outside of that, the limit will lie within that range. Thus,
$$\lim_{r \to \infty} \arg{(r e^{i \theta}+z_0)} = \theta - \bigg \lceil \frac{\theta}{2 \pi} \bigg \rceil 2 \pi $$
Ron Gordon
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