This may be a strange question; but I've read and re-read the chapter in my textbook on what exactly a branch of a logarithm is and am having trouble understanding.
What is a branch of a logarithmic function? And what is a branch cut- how does it relate to the analyticity of a function?
Imagine trying to define the square root function $f(x) := x^{1/2}$ on the unit circle.
Start with $x = 1 = e^{2\pi i\cdot 0}$, then you have two choices for $f(1)$. Say you decide that $f(1) = 1$. Alright, but you also want $f(x)$ to be continuous, so as $x$ moves counterclockwise along the unit circle starting at 0, $f(x)$ should also move continuously. In this case, the function you're basically looking at is $$f(x) = e^{2\pi i t/2}$$ where $t$ is a real variable and $x = e^{2\pi it}$. As $t$ increases, $x$ will move along the unit circle counterclockwise, and $f(x)$ will also move along the unit circle counterclockwise, but at "half the speed" that $x$ does. The problem occurs when $t = 1$ corresponding to $x = 1$. By the original definition, $f(1) = 1$, but at $t = 1$, $e^{2\pi i t/2} = -1$, which shows that your attempt to define such a continuous square-root function via the formula $f(x) = f(e^{2\pi it}) = e^{2\pi it/2}$ fails.
Of course, this doesn't prove anything, but from this you should be able to see why you can't define a square root function on the unit circle, so you certainly can't define on $\mathbb{C}$ or $\mathbb{C}^\times$. However, if you remove a point $b$ from the unit circle $S^1$, you can define a square root function on $S^1\setminus \{b\}$, since the problem described above no longer occurs.
Removing the point $b$ is analogous to taking a branch cut.
But this is the square root function! Not the log! Well, perhaps this observation will help:
$$e^{\frac{1}{2}\ln x} = e^{\ln x^{1/2}} = x^{1/2}$$ Thus, if $\ln$ was holomorphic and definable on $\mathbb{C}^\times$, then so must $x^{1/2}$.
