I have some questions about knowing where and where not functions are analytic. Here's a function,
f(z)= $\frac{Log(z+4)}{z^2+i}$
-I know that this function is not defined for z=$\pm$$\frac{(1-i)^2}{\sqrt2}$ being as though these are the solutions that make the denominator equal to 0. Why would it also not be analytic on the x-axis for x$\leq$-4?
On
Actually, to maximise the set on which $f$ is analytic, chose the branch cut of $\log$ to travel through one of the zeros of the denominator. $$f: \mathbb C \setminus (\{z_1\} \cup \{\lambda z_2 -4; \mathbb R \ni \lambda \geq 0\})$$ Where $z_{1,2}$ are the zeros of the denominator is the maximum possible area of analyticity.
However if you want to stick to the principal branch of the $\log$ it is defined on $\mathbb C \setminus (-\infty, 0]$, but you can cut any line from $0$ to any direction of $\infty$. Ideally you cut somewhere, where you already lost definition domain by the denominator, which is what I suggested. But if you want to use the principal $\log$, $$\log: \mathbb C \setminus (-\infty, 0] \to \mathbb C$$ Then $\log (\cdot + 4) : \mathbb C \setminus (-\infty, -4] \to \mathbb C$ So this is why the analyticity goes away there.
That is exactly correct. The principal logarithm is usually defined to have a branch cut along the negative real axis, so $Log(z)$ is undefined for $z\in\mathbb{R}^-$.