Why is branch cut needed in complex log function? (I asked this in chat but did not get helpful reply)
This is from Kreyszig Advanced Engineering Mathematics:


Where formula (3) is $\ln z = \ln |z| + i \text{ Arg } z \pm 2n\pi i$

I can't understand what would happen if, say for $n=0$, we include negative real axis as well? I mean, derivative is obtained to be $1/z$ even at negative real axis, hence in particular it is continuous there, right?  (using 'a function differentiable at a point is continuous there')


That is, Why instead of working with $\text{Ln } z = \ln |z| + i \text{ Arg } z$ ($-\pi<\text{Arg } z\le \pi$) (which is indeed a function in conventional sense), this theorem deals with, say, $\ln |z| + i \text{ Arg } z$ for $-\pi<\text{Arg } z< \pi$?

 A: It is impossible to create a continuous extension of the complex logarithm to $\mathbb{C}-\{0\}$. If the function is not continuous, it cannot possibly be differentiable. The necessary conclusion is that any supposed proof of differentiability contains an error.
Why is it impossible to create a continuous extension of logarithm to the punctured plane? By contradiction, suppose we had a function $L:\mathbb{C}-\{0\}\rightarrow \mathbb{C}$ such that $L(z)$ is complex differentiable to $\frac{1}{z}$ everywhere except $z=0$. Now compose our function with the complex exponential, and apply chain rule.
$$\frac{d}{dz}L(e^z)=\frac{1}{e^z}\cdot \frac{d}{dz}e^z=e^{-z}e^{z}=1$$
So $L(e^z)$ has derivative $1$ everywhere. Note that this applies for all $z$, since $e^z$ is never zero, and we had assumed that $L$ was differentiable everywhere except at $0$. Now integrate to reverse the differentiation.
$$L(e^s)-L(e^z)=\int_{z}^{s}1dz=s-z$$
Let $z=0$ and let $s=i2\pi$, and while recalling that $e^{i2\pi}=1$, we get:
$$L(e^{i2\pi})-L(e^0)=i2\pi-0 \implies L(1)-L(1)=i2\pi \implies 0=i2\pi$$
A contradiction. So the function $L$ with the described properties cannot exist.
In case you are wondering which error was made in differentiation: It's arctan. Logarithm needs a branch cut because arctan needs a branch cut. You can put the branch anywhere you want, and on any point other than the cut, it will be differentiable, but you need to put it somewhere, and wherever you put it won't be differentiable. For the usual branch, arctan(y/x) is discontinuous whenever $x=0$, regardless of $y$.
