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Be $z\in\mathbb{C}\setminus\{-1\}$ .

Condition: $\enspace\displaystyle (1+\frac{1}{z})^{z+1}=\overline{(1+\frac{1}{z})^z}$

Wanted: $\enspace$ The complete solution set.

Note:

Be $\enspace\displaystyle e^{i\alpha}:=\frac{\overline{(1+\frac{1}{z})^z}}{(1+\frac{1}{z})^z}$ ,$\enspace\alpha\in\mathbb{R}$ .

It's easily to see that $\displaystyle 1+\frac{1}{z}=e^{i\alpha}$ which means

$\displaystyle z=\frac{1}{e^{i\alpha}-1}=-\frac{1}{2}-\frac{i}{2}\cot\frac{\alpha}{2}\enspace $ and therefore always $\enspace\displaystyle \Re(z)=-\frac{1}{2}$ .

$z\enspace$ is here only valid, if $\enspace\alpha\in\mathbb{R}\setminus 2\pi\mathbb{Z}$ .

Which solutions are still missing (other cases, $\displaystyle \Re(z)\neq-\frac{1}{2}$) ?

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    You have found a necessary condition, but not a sufficient one... – 2017-02-15
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    @b00nheT : I *set* it, not *found* . --- I am looking for all solutions, not only for one. :-) – 2017-02-15
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    What I am saying, is that so far you have found **zero** solutions, so your sentence "which solutions are still missing" is somewhat funny – 2017-02-15
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    @b00nheT : Would be nice if you tell me more about the problem with the word *solutions*, maybe I can correct my sentence so that other interested people can understand better what I mean. :-) – 2017-02-15
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    Right now you have shown that **if** z is a solution, then $\Re(z) = -\frac{1}{2}$. That doesn't mean that the converse holds, i.e. all $z$ with $\Re(z) = -\frac 12$ are indeed solutions. This is what @b00nheT means when he says that you have no solution so far because it is possible that no $z$ with $\Re(z)= -\frac 12$ is a solution. I am also not so sure about that $\cot$ thing, so i don't know if what you have shown is correct since I'm not familar with those functions on coomplex domains but this might be true, i dont know. – 2017-02-15
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    @LeBtz : Thanks(!) for your explanation, I should be more precise. I have specified the valid range. --- The $\cot$ ? It's only $\frac{1}{e^{i2x}-1}=\frac{e^{-ix}}{i2\sin x}=...$ . – 2017-02-15

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