I am relatively new to complex analysis, and in my course we have to calculate the following limits (if they exist):
$\lim \limits_{z \to 1}\frac{log(z)}{z}~~~~~$ $\lim \limits_{z \to -1}\frac{log(z)}{z}~~~~~$ $\lim \limits_{z \to 0} z \cdot log(z)$
The first one should be clear, since log is defined at $1$ and is continuous in the domain $\mathbb{C}\backslash\mathbb{R}_{\leq 0}$. So I can simply evaluate $\lim \limits_{z \to 1}\frac{log(z)}{z} = \frac{log(1)}{1} = 0$
right ?
I am not sure about the second and third. I know that L'Hospitals rule works in the complex numbers as well, so i am supposed to get $\lim \limits_{z \to -1}\frac{log(z)}{z} = -1$ and
$\lim \limits_{z \to 0} z \cdot log(z) = 0$. At least if I am not forgetting about some precondition of L'Hospital. But this is not what I am supposed to do, since I haven't even gotten introduced to complex derivations yet. Is there a simple way yo evaluate these limits if I only have elementary definitions of complex analysis?
$\lim \limits_{z \to -1}\frac{log(z)}{z}~~~~~$ for this limit, I would use the fact since $i^2=-1$ and by log rules change $\frac{log(-1)}{-1}$ to $\frac{log(i)}{-1}+\frac{log(i)}{-1}$
As @joséCarlosSantos stated $log(z)=log(||)+iarg(z)$ this leads to $log(i)=\frac{i}{2}\pi$ thus:
$\lim \limits_{z \to -1}\frac{log(z)}{z}$=$\frac{log(-1)}{-1}$ =$\frac{log(i)}{-1}+\frac{log(i)}{-1}$=$\frac{-i}{2}\pi$$\frac{i}{-2}\pi$=$-i\pi$