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 $\mathrm{Arg}(z)$.

  • $\begingroup$ That's really simple and clear; thank you! Is this also assuming continuity of the "Arg" function? $\endgroup$ – Twenty-six colours Jul 22 '17 at 10:53
  • 1
    $\begingroup$ It does not assume anything. You can prove it starting by the relation between Polar and Cartesian coordinates. But it does shows that $\text{Arg}$ function is continuous everywhere, except on the negative part of the real axis. $\endgroup$ – Naive Jul 23 '17 at 5:03

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 $[0,2\pi)$.


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.

  • $\begingroup$ Hmm, I'm not sure how I would have proven it without knowing the result beforehand. I'm not sure what it means by "approaching from above" still when the imaginary part is non-negative. Do I take a path $y = -t$ for $-\infty < t < \infty$ or something? $\endgroup$ – Twenty-six colours Jul 19 '17 at 14:13
  • $\begingroup$ If you let $z=re^{i\theta}$ then the path you take to reach the negative axis "from above" (the real axis) is $0$ to $\pi$. The path you take to reach the negative axis "from below" (the real axis) is $0$ to $-\pi$. See my update. $\endgroup$ – Pixel Jul 19 '17 at 15:47

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