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I have an observation, and I don't know that the following statement is true or not. If not give a counterexample, if it is true prove it, or give a reference about it.

Let $n \in \mathbb{R}$, $z \in \mathbb{C}$ and denote the polylogarithm function with $\operatorname{Li}_n$.

$(a)$ If $\Im z \neq 0$, then $\Im \operatorname{Li}_n(z) + \Im \operatorname{Li}_n\left({\overline z}\right) = 0,$

$(b)$ $\Re \operatorname{Li}_n(z) - \Re \operatorname{Li}_n\left({\overline z}\right) = 0,$

where $\Im$ denotes the imaginary part of a complex number, $\Re$ denotes the real part of a complex number and ${\overline z}$ denotes the complex conjugate of $z$.

If you can tell us something just about special cases you're also welcome. The most preferred and interesting case for me is $\operatorname{Li}_3$.

user153012
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1 Answers1

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$\def\Li{{\rm Li}}$Note that generally $\Li_n$ is single valued analytic function in $D=\mathbb{C}\setminus[1,+\infty)$. This domain is symmetric with respect to the real axis. Now, $g(z)=\overline{\Li_n(\overline{z})}$ is also analytic in the same domain $D$ and it coincides with $\Li_n$ on the real interval $(-1,1)$. This implies that $g=\Li_n$ on $D$. $$ \forall\,z\in D,\qquad \Li_n(z)=\overline{\Li_n(\overline{z})} $$ From this we conclude that $$\eqalign{ \Re(\Li_n(z))&=\frac{1}{2}\left(\Li_n(z)+\overline{\Li_n(z)}\right)= \frac{1}{2}\left(\Li_n(z)+\Li_n(\overline{z})\right)\cr \Im(\Li_n(z))&=\frac{1}{2i}\left(\Li_n(z)-\overline{\Li_n(z)}\right)= \frac{1}{2i}\left(\Li_n(z)-\Li_n(\overline{z})\right)\cr } $$ In particular, for $z\in D$, we have $$\eqalign{ \Li_n(z)+\Li_n(\overline{z})&\in\mathbb{R}\cr \Li_n(z)-\Li_n(\overline{z})&\in \mathbb{R} i } $$ from which the desired conclusion follows.

Omran Kouba
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  • Prof, what does this statement mean: "$\text{Li}_n$ is single valued analytic function in $D=\mathbb{C}\setminus[1,+\infty)$"? What are "single valued analytic function" and "$\mathbb{C}\setminus[1,+\infty)$"? – Anastasiya-Romanova 秀 Sep 20 '14 at 14:58
  • @Anastasiya-Romanova, It is like the complex logarithm, the logarithmic function $\ln$ has a unique analytic continuation to $\mathbb{C}\setminus(-\infty,0]$, It cannot be extended analytically any further while remaining analytic and single valued, if we try to do so we obtain a "multiply valued function", May be you would like to read this: http://www.maths.dur.ac.uk/Ug/projects/highlights/PR4/Rhodes_Polylogarithms_report.pdf. – Omran Kouba Sep 20 '14 at 15:21
  • Thank you Prof for replying my comment & also for providing me a paper to learn. +1 (^◡^ ) – Anastasiya-Romanova 秀 Sep 20 '14 at 19:40