Using the Cauchy-Riemann Equations, prove that $\operatorname{Log}(z)$ is holomorphic on $\mathbb{C}\setminus[0,\infty)$ and explain why is it only holomorphic in that region.
In our course, $\operatorname{Log}(z)$ with upper-case L refers to the principal branch of log where $0\leq Arg(z)<2\pi$
My attempt:
$$\operatorname{Log}(z)=\ln|z|+i\operatorname{Arg}(z)$$
Let $r:=|z|=\sqrt{(x^2+y^2)}$ and let $\theta:=\operatorname{Arg}(z)=\arctan(\frac{y}{x})$
And thus
$$\operatorname{Log}(z)=\ln(r)+i\theta$$
Let $u=\ln(r)$ and $v=\theta$. For the C.R.Es to be satisfied:
$$u_x=v_y$$ and $$u_y=-v_x$$
must hold for some region.
$u_x=\frac{du}{dr}\frac{dr}{dx}$ $u_y=\frac{du}{dr}\frac{dr}{dy}$
$v_x=\frac{dv}{d\theta}\frac{d\theta}{dx}$ $v_x=\frac{dv}{d\theta}\frac{d\theta}{dx}$
$$u_x=\frac{1}{r}\frac{x}{r}=\frac{x}{r^2}$$ $$u_y=\frac{1}{r}\frac{y}{r}=\frac{y}{r^2}$$
$$v_x=\frac{-y}{r^2}$$ $$v_y=\frac{x}{r^2}$$
So for some region, $\operatorname{Log}(z)$ has partial derivatives that satisfy the C.R.Es. Meaning it is holomorphic in that region.
As for the reason behind the region. I feel it is related to $\arctan$ being continuous only in that region. Can someone explain this part? Also, graphically, where is this region?
In the usual definition of principal logarithm the argument is taken to be in $(-\pi, \pi]$. With this definition the correct domain of analyticity is $\mathbb C \setminus (-\infty, 0]$. So the statement requires a modification. [Of course if you redefine Log using argument of $(0,2\pi]$ then the statement is correct]. Your argument for the first part is correct. Hint for showing that Log is not continuous on the negative real axis: show that $\arg(-x+\frac 1 n) \to \pi$ and $\arg(-x-\frac 1 n)\to -\pi$ for any $x>0$.