Let $x\in [a,b]$ with $a<b$ real numbers.
Let Log be the principle determination of the complex logarithm with argument in $]-\pi,\pi[$.
I try to calculate $\lim_{z\to x, Im(z)>0} f(z)= Log(\frac{z-a}{z-b})$.
What I did $f$ is well defined on $ [a,b]$. $Log(\frac{z-a}{z-b}) = |\frac{z-a}{z-b}| + i Arg(\frac{z-a}{z-b})$.
The modulus and the argument are continuous functions so:
$\lim_{z\to x, Im(z)>0} f(z)=-\frac{x-a}{x-b} + i Arg(\frac{x-a}{x-b})$
My question is about $Arg(\frac{x-a}{x-b})$, is it simply $\pi$ because it it a negative real?
Thank you for your help.