I recently came across this question and I posted an answer. It has been pointed out that my answer is incorrect. I cannot work out what is wrong with my reasoning. The answer I gave corresponds with the Abel and Cesaro sum, so perhaps $\sum$ is not the usual summation operator? Am I correct in asserting that if $x$ is in the upper half-plane, i.e., $\textbf{I}[x] > 0$, then $|e^{ix}| < 1$ and consequently $$\sum_{n = 1}^\infty e^{inx} = \frac{e^{ix}}{1 - e^{ix}},$$ or is my argument flawed? Any help would be appreciated.
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Ah, no, the problem is when you assume $1-e^{-ix}$ is the conjugate of $1-e^{ix}$. It is only the conjugate when $x$ is real... – Thomas Andrews Jun 28 '14 at 23:44
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You never have $|e^{ix}|<1$. Because $|e^{ix}|=1$ for every $x$. – EQJ Jun 28 '14 at 23:45
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Also, the imaginary part of $e^{inx}$ is not $\sin nx$ when $x$ is not real. – Thomas Andrews Jun 28 '14 at 23:45
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@Yotas follow the link, he assumes the imaginary part of $x$ is positive... – Thomas Andrews Jun 28 '14 at 23:46
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@YotasTrejos You are wrong. Like I said, $|e^{ix}| < 1$ when $x$ is in the upper half-plane. – glebovg Jun 28 '14 at 23:47
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@ThomasAndrews I think you are right. – glebovg Jun 28 '14 at 23:47
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Because you assume that $x$ is not real, the imaginary part of $e^{inx}$ is not $\sin nx$.
Also, when computing the conjugate if $1-e^{ix}$, you don't get $1-e^{-ix}$ when $x$ is non-real, but rather $$1-e^{-i\bar x}$$
Thomas Andrews
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I know that $1 - e^{-ix}$ is not the conjugate of $1 - e^{ix}$. I simply multiplied and divided by it to find the real and imaginary part of $e^{ix}/(1 - e^{ix})$. Is this not permissible? – glebovg Jun 28 '14 at 23:53
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That's wrong, because the denominator needs to be real for that trick to work, and the only way for that trick to work is if you multiply by the conjugate. $(1-e^{ix})(1-e^{-ix})$ is not real when $x$ is not real, so how does that help you? – Thomas Andrews Jun 28 '14 at 23:55
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I see. You are right. I think this answers my question. However, I find it rather strange that my reasoning lead to the correct Abel and/or Cesaro sum. – glebovg Jun 28 '14 at 23:56
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I have a feeling about the Abel summability of a general power series on its boundary when the function has an analytic continuation on that boundary point. But that's just a gut feeling. – Thomas Andrews Jun 29 '14 at 00:13
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Thanks again. By the way, I have a feeling that there is not closed formula for the real and imaginary part of $e^{ix}/(1 - e^{ix})$ when $x$ is in the upper half-plane. Am I correct? – glebovg Jun 29 '14 at 00:38