Confusion regarding pole

Question

Consider the following complex-valued function. $$f(z)=\frac{\log(z+i)}{1+z^2}$$. The book which i am referring to states that at $z=-i$, the function has a logarithmic singularity.However , $$\lim_{z\to -i}f(z)=\infty$$ by the L'hospital's rule which suggests that at $z=-i$ , the function should have a pole. Can someone point out where i am wrong ? Thanks in advance.

Answer 1

The use of l’Hospital’s rule is not justified. The denominator $1 + z^2$ does evaluate to $0$ when $z = -i$, but the numerator $\log(z + i)$ doesn’t (undefined). The numerator doesn’t even approach $0$ as $z\to -i$. In fact, its real part approaches $-\infty$ no matter which branch of $\log$ is used.

In any case, the singularity of $\log(z + i)\big/\big(1 + z^2\big)$ at $z = -i$ is not isolated. So, even if this limit is $\infty$, it has no bearing on whether it is a pole, because it’s not an isolated singularity. One requirement in the definition for a pole is that it be an isolated singularity.