Show that $\operatorname{Log}\left ( 1+\frac{1}{z} \right )=\operatorname{Log}(z+1)-\operatorname{Log}(z)$ It is well-known that whenever $a,b$ are non-zero complex numbers, then $\operatorname{Log}(a/b)=\operatorname{Log}(a)-\operatorname{Log}(b)+2\pi i k$ for some integer $k$. Here $\operatorname{Log}$ denotes the principal logarithm. I read somewhere without explanation that
$$
\operatorname{Log}\left ( 1+\frac{1}{z} \right )=\operatorname{Log}(z+1)-\operatorname{Log}(z)\tag{*}
$$
for all $z\in \mathbb{C}\setminus (-\infty,0]$. Is this valid in general? If so, why? I checked it myself to insert some values of $z$ in $\mathbb{C}\setminus (-\infty,0]$ and compare on both sides, and they look good. However, I can't seem to prove it in general, or that $k$ should be $0$ in that case (where $a=z+1$ and $b=z$).
Update: I have tried a slightly different way, though I might err:
Write $x=a+ib$, then
$$
\frac{z+1}{z}=1+\frac{a}{a^2+b^2}+i\frac{-b}{a^2+b^2}.\tag{**}
$$
If $a>0$, (*) is valid, as equality holds whenever $\Re z+1,\Re z>0$.

*

*If $a=0$, then (**) is reduced to $1-i/b$, so $\operatorname{Log}(\frac{z+1}{z})=\ln|1-i/b|+\arctan(-1/b)$, while $\operatorname{Log}(z+1)-\operatorname{Log}(z)=\ln|1-i/b|+\arctan(b)-\operatorname{Arg}(ib)$.


*$\qquad$ If $b>0$, then the latter would be equal to $\ln|1-i/b|+\arctan(b)-\pi/2$.


*$\qquad$ For $b<0$, we would instead get $\ln|1-i/b|+\arctan(b)+\pi/2$.
In both cases, I suspect that $\arctan(-1/b)$ is equal to $\arctan(b)-\pi/2$ for $b>0$ and equal to $\arctan(b)+\pi/2$ for $b<0$. I checked it with Graph calculator, and left hand side and right hand side seem to coincide. How to prove it rigorously? See update 2 below.
If $a<0$, replace $a$ by $-a'$ for $a'>0$, then (*) is still valid for $a'>0$.
Update 2: Define $f(x)=\arctan(-1/x)-\arctan(x)$ for $x\neq 0$. Then, we see that $f'(x)=0$, so $f$ must surely be constant. But, since $f(1)=-\pi/2$, it follows that $f(x)=-\pi/2$ for all $x\neq 0$. But, this is not entirely true, as $f(x)$ is $\pi/2$ for negative $x$. Where is the flaw in my argument?
 A: It is not difficult to see that, in general, if $w_1,w_2 \in \mathbb{C}\setminus (-\infty,0]$ then
$\operatorname{arg} {w_1 \over w_2} = \operatorname{arg} w_1 - \operatorname{arg} w_2 + 2k \pi$, where $k \in \{-1,0,+1\}$. So, the goal here is to show that $k=0$.
Split the analysis into three cases depending on the sign of $\operatorname{im} z$. Let $\operatorname{arg}w$ denote the angle of $w$ in $(-\pi,\pi)$.
Suppose $\operatorname{im} z > 0$, then $0< \operatorname{arg} (1+z) < \operatorname{arg} z < \pi$ and hence
$\operatorname{arg} (1+z) - \operatorname{arg} z \in (-\pi,0]$.
Note that ${1+z \over z} = {\bar{z} + |z|^2 \over |z|^2}$, from which we see that $\operatorname{arg} {1+z \over z} \in (-\pi,0]$, and so $k=0$.
The other cases, $\operatorname{im} z = 0$ and $\operatorname{im} z <0 $ are similar.
A: If you draw a picture, the geometry of this problem becomes immediately clear. In general, the property that makes this work is the fact that $z$ and $z+1$ will always have arguments who's difference is less than $\pi$. That is, you can't have $z$ and $z+1$ in QII and QIII at the same time, they cannot be vertical from each other. This would  fail for any non-real number, for example,
$$\text{Log}\left(\frac{z+i}{z}\right) \not= \text{Log}(z+i) - \text{Log}(z)$$
at $z=-1-\frac{1}{2}i$. Likewise, if we chose a different branch of the logarithm that wasn't on the real line, then original statement with a $1$ would now fail too. Again, a picture really makes this clear.
