Showing $\operatorname{Log}(z-i)$ is not analytic Show that the function $\operatorname{Log}(z-i)$ is analytic everywhere except on the half line $y=1$ $(x\leq 0)$. 
I know that $$\operatorname{Log}(z-i)=\operatorname{Log}(x+iy-i)=\operatorname{Log}(x+i(y-1))=\ln\sqrt{x^2+(y-1)^2}+i \operatorname{Arg}(x+i(y-1))$$ I get that this function cannot be differentiated at $x=0, y=1$ due to the fact $\ln(0)$ is undefined. Is it because of this reason it is not analytic everywhere else on that line? 
 A: No, the problem is not the real logarithm, but the argument. Now I explain.
The complex logarithm, called principal logarithm, is defined as follows:
$\log_{\mathbb C}z:=log_{\mathbb R}|z|+i\arg z\;\;,$ where $\arg z=$
$\tan^{-1}\left(\frac{\Im z}{\Re z}\right)$ if $\Re z>0$
$\cot^{-1}\left(\frac{\Re z}{\Im z}\right)\;\;\;\;$ if $\Re z\leq0$ and $\Im z>0$
$\cot^{-1}\left(\frac{\Re z}{\Im z}\right)-\pi\;\;\;\;\;$ if $\Re z\leq0$ and $\Im z<0$
Now this definition could seem complicated, however you see that the only problem is that is not defined in the half line $\{z\in\mathbb C\;\;:\;\;\Im z=0,\; \Re z\leq0\}$. But 
$$
\lim_{\Im z\rightarrow0^+,\Re z<0} \cot^{-1}\left(\frac{\Re z}{\Im z}\right)=
\lim_{t\rightarrow-\infty}\cot^{-1}(t)=\pi
$$
$$
\lim_{\Im z\rightarrow0^-,\Re z<0} \cot^{-1}\left(\frac{\Re z}{\Im z}\right)-\pi=
\lim_{t\rightarrow+\infty}\cot^{-1}(t)-\pi=0-\pi=-\pi
$$
So the best you can do is to extend the function on this line (by posing $\arg_{|\mathbb R_{<0}}\equiv\pi$ for example)... but however it won't be continous here! 
Now the argument of the complex number $z$ is to be thought as the angle that your complex number $z$ (thought as a vector of $\mathbb R^2$) forms with the real line axis; so it's a number $\theta\in]-\pi,\pi[$.
Now it should be clear that the complex logarithm couldn't be continous where the real log is not (only in $0$, since you consider $|z|$, that is $\geq0$) and where the argument is not (in the whole $\mathbb R_{\leq0}$, as explained). Otherwise is continous hence analytic. So you can desume that $\log_{\mathbb C}\in \mathcal{H}(\mathbb C_-)$ where $\mathbb C_-:=\mathbb C\setminus\mathbb R_{\leq0}$.
Hence considering 
$\log_{\mathbb C}(z-i)$ you see immediately that the "critic half line" is parallel to the one of $\log_{\mathbb C}z$ but passing by $i$. Hence arguing exactly as already done, you can conclude that 
$\log_{\mathbb C}(z-i)$ is analytic on $\mathbb C\setminus\{z\in\mathbb C\;\;:\;\;\Re z\leq0,\;\Im z=1 \}$, as you said.
