Suppose that $a>0$. Show that
$$\lim_{\substack{z\rightarrow -a \\ \Im(z)\geq 0 }}\mathrm{Arg}(z) = \pi$$
and
$$\lim_{\substack{z\rightarrow -a \\ \Im(z) < 0}}\mathrm{Arg}(z) = -\pi.$$
I'm not sure how to approach this question. I don't understand the significance of the (for example) first limit having a non-negative imaginary part for $z$. Does this mean it's approaching "from above" or something?
For an arbitrary $z\neq0$, the principal value of $\arg z$ is defined to be the unique value that satisfies $-\pi<\arg z\le\pi$ (well we can define it over any interval of length $2\pi$, in your case we stick to the interval $(-\pi,\pi ]$) and it is usually denoted by $\mathrm{Arg}(z)$. Thus the relation between $\arg z$ and $\mathrm{Arg}(z)$ is given by
$$\arg z=\mathrm{Arg}(z) +2k\pi,\;\;\;k\in\Bbb Z$$
You can deduce the following starting from the relation between Cartesian and the Polar form,
$$\mathrm{Arg}(z) = \begin{cases} \tan^{-1}\left(y/x \right), & \text{if $x>0$} \\[2ex] \pi+\tan^{-1}\left(y/x \right), & \text{if $x<0,y\ge0$} \\[2ex] -\pi+\tan^{-1}\left(y/x \right), & \text{if $x<0,y<0$} \\[2ex] \pi/2 , & \text{if $x=0,y>0$} \\[2ex] -\pi/2 , & \text{if $x=0,y<0$} \end{cases}$$
That's all you need to know before starting to solve the problem.
====================================================================
Let's start with showing, $\lim_{\substack{z\rightarrow -a \\ \Im(z)\geq 0 }}\mathrm{Arg}(z) = \pi$
Here $-a$ lies on the negative part of the real axis.
Let $z=x+iy\;\;,x,y\in\Bbb R$
As $z\to -a$ with the condition $\Im(z)\geq 0 \implies x\to -a,y\to 0$ with $y\ge 0$, that is we approach $-a$ from the upper half of the plane.
Hence by second case among values of $\mathrm{Arg}(z)$,
$$\lim_{\substack{z\rightarrow -a \\ \Im(z)\geq 0 }}\mathrm{Arg}(z) = \lim_{\substack{(x,y)\rightarrow (-a,0) \\ y\geq 0 }}\mathrm{Arg}(x,y) = \lim_{\substack{(x,y)\rightarrow (-a,0) \\ y\geq 0 }}\pi+\tan^{-1}\left(y/x \right)=\pi$$
Secondly for showing, $\lim_{\substack{z\rightarrow -a \\ \Im(z) < 0}}\mathrm{Arg}(z) = -\pi$, you can follow similar arguments but in this case we have $\Im(z) < 0$. So as $z\to-a$, we approach $-a$ from the lower half of the plane, that is with $y<0$. Hence we use the third case among the values of $\mathrm{Arg}(z)$.
Yes it's to do with the approach. If $a>0$ then $a$ is a real number, so that $-a$ negative and real. Hence if $z\to-a$ from above (i.e. $\Im z\geq 0$) then $z$ tends to a negative real number from above and the argument is $\pi$. On the other hand if $z\to-a$ from below (i.e. $\Im z<0$) then $z$ tends to a negative real number from below and the argument is $-\pi$. Of course this all depends on how you define your Principal Argument ($\text{Arg}$). In your question they are clearly using $[-\pi,\pi)$, but you could also use $[0,2\pi)$.
UPDATE
Loosely speaking, let $z=r(\cos\theta+i\sin\theta)=re^{i\theta}$ and $r\neq 0$. Assume we use $[-\pi,\pi)$ for our principal argument. If $\Im z\geq 0$ then it must be that $0\leq\theta<\pi$, so the complex number is "above" the real axis. Then $\lim z\to-a$ from above is the same as $\theta\to\pi$. If $\Im z<0$ then it must be that $-\pi\leq\theta<0$, which is "below" the real axis. Then $\lim z\to-a$ from below is the same as $\theta\to-\pi$. In either case the complex number is approaching the negative number line from above or below the real axis.
