TITLE: Find $\lim\limits_{y\to 0}\,\log(a+iy)-\log(a-iy)$ with principal value of $\log$
QUESTION [0 upvotes]: So I'm working with the principal value of the complex logarithm. Let $U$ be the plane with the negative real axis deleted. Let $y>0$. Find the limit
$$\lim_{y\to 0} [\log(a+iy)-\log(a-iy)].$$
for $a>0$, but also for $a<0$. 
So for $a>0$ it is not that hard I guess, since 
\begin{align*}
\log(a+iy) &= \log\sqrt{a^2+y^2} + i\arg(a+iy)
\end{align*}
and 
\begin{align*}
\log(a-iy) &= \log\sqrt{a^2+y^2} + i\arg(a-iy) \\
&= \log\sqrt{a^2+y^2} - i\arg(a+iy)
\end{align*}
so 
$\log(a+iy)-\log(a-iy) = 2i\arg(a+iy)$. 
Please tell me if I'm making wrong conclusions. Now if $y\to 0$, $2i\arg(a+iy)$ will go to 0 and that's the limit. 
Is this the correct way of thinking? And how to proceed with $a<0$? 
Thanks!

REPLY [1 votes]: This is correct, though it also suffices to notice that away from a branch cut $\log$ is continuous, so for $a > 0$, $\lim_{y \to 0} \log(a + iy) = \log a = \lim_{y \to 0} \log(a - iy)$ and hence $$\lim_{y \to 0} [\log(a + iy) - \log(a - iy)] = 0 .$$
For $a < 0$ we can't use this method, as the limit $\lim_{y \to 0} (a \pm iy)$ lies on the branch cut, so $\log$ isn't continuous there. There are a few ways to handle the problem from here. A straightforward way is to use the explicit formulas for the standard branch of $\arg$ in the second and third quadrants; for the (open) second quadrant, this is $\arg(x + i y) = \arctan \frac{y}{x} + \pi$, so
$$\lim_{y \searrow 0} \arg(a + iy) = \lim_{y \searrow 0} \left(\arctan \frac{y}{a} + \pi \right) ,$$
but the quantity in parentheses is a continuous function of $y$, so the limit is given just by evaluating and thus
$$\lim_{y \searrow 0} \arg(a + iy) = \pi .$$
Now, treat this limit in the direction $y \nearrow 0$, as well as the limits of $\arg(a - iy)$, and compute.
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