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 $D_0= R_0 \setminus D_0$

  • $\begingroup$ 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

$\text{Log}$ 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 $$\eqalign{f(5 i) &= \ln(5) + i \pi/2\cr f(-6) &= \ln(6) + i \pi\cr f(-7i) &= \ln(7)+ 3 i \pi/2\cr f(8) &= \ln(8) + 2 i \pi\cr f(9i) &= \ln(9) + 5 i \pi/2\cr f(-10) &= \ln(10) + 3 i \pi\cr &etc.}$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.