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Let $ U ⊂ \mathbb{C}$ be a simply-connected region, and suppose $\exp(f(z))=z ∈ U$. Then there is a unique analytic function $f(z) : U → C$ such that $f(z) = \operatorname{Log} (z)= \ln|z|+i \operatorname{Arg}(z)$ with $f(-1)=+i\pi $ for all $z ∈ U$ with $0<\operatorname{Arg}(z) \leq 2\pi$.

If we take a closed set $L_{0},$ such that, $L_{0}= \{1+iy : -\infty <y\leq 0\}.~$ There is a unique analytic function in $~ R=U \setminus L_{0},~$ such that $f(z)= \ln|z-1|+i \operatorname{Arg}(z-1),~$ and satisfy that $exp(f(z))= z-1,~$ with $f(-2)=\ln|-2|+i\pi,~ $ for all $z ∈ R,~ $ with $\frac{- \pi}{2}<\operatorname{Arg}(z-1) \leq \frac{3 \pi}{2}$.

My question, is it tru to write that $f(z)= \operatorname{Log}(z-1)+ 2 \pi i k ~$ for all $z$ in the domain $D_{k}$ and $k=0,1$ such that $~D_0=\{z \in \mathbb{C} : \operatorname{Re}(z)>1 , \operatorname{Im}(z)\leq 0 \}~$ and $~D_1= R_0 \setminus D_0$

Or $f(z)= \operatorname{Log}(z-1)+ 2 \pi i k +\pi i$ for all $z$ in the domain $D_{k}$ and $k=-1,0 $ such that $D_{-1}=\{z \in \mathbb{C} : \operatorname{Re}(z)>1 , \operatorname{Im}(z)\leq 0 \}~$ and $D_0= R_0 \setminus D_0$

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  • $\begingroup$ If you have a followup to someone else's answer, you should not post that followup as an answer to your question. Instead you can post your followup as a comment to that person's answer. Or, you can edit your question by adding something like "The answer of Joe Schmo leads me to ask a refined question ..." $\endgroup$ – Lee Mosher Jan 29 '16 at 21:36
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$\text{Log}$ is usually used for the principal branch of the logarithm, where the argument is in the interval $(-\pi,\pi)$. That has branch cut on the negative real axis, and if $U$ intersects that branch cut your $f(z)$ is certainly not $\text{Log}(z)$. Indeed, you could have a simply connected region like this:

enter image description here

where the imaginary part of $f(z)$ is forced to have an arbitrarily large range. As you go around the spiral counterclockwise, the argument increases by $2 \pi$ on every turn.

EDIT: If $f(4) = \ln(4)$, then $$\eqalign{f(5 i) &= \ln(5) + i \pi/2\cr f(-6) &= \ln(6) + i \pi\cr f(-7i) &= \ln(7)+ 3 i \pi/2\cr f(8) &= \ln(8) + 2 i \pi\cr f(9i) &= \ln(9) + 5 i \pi/2\cr f(-10) &= \ln(10) + 3 i \pi\cr &etc.}$$

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  • $\begingroup$ Beautiful picture, that! Maybe you could write down the values of $f(-5i), f(6), f(15i), f(-17)$ for the sake of explicitness ? $\endgroup$ – Georges Elencwajg Jan 29 '16 at 20:26
  • $\begingroup$ What about my question if take argument such that [0,2π)? $\endgroup$ – Mfs Jan 29 '16 at 22:02
  • $\begingroup$ It doesn't. The more turns of the spiral, the bigger an interval you need. $\endgroup$ – Robert Israel Jan 29 '16 at 22:12

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