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I am reading Complex Analysis by David C. Ulrich. In appendix 3 titled "Sin, Cos and Exp", Ulrich defines $\exp , \sin , \cos$ by using the power series. After a few lines, he writes that

The absolute convergence of the series shows that the terms can be rearranged as desired, and it follows that \begin{align} \exp (iz) = \cos (z) + i \sin (z) \end{align}

But it looks to me that rearrangement is not really necessary from this question which I asked on this site. Although it is not necessary, how can it be done? I have tried a lot but I do not see it. Hints are appreciated!

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  • $\begingroup$ Compute the first several terms for exp(ix), simplify them. Some will be real, others complex; separate them like this $\endgroup$
    – David Raveh
    Jan 8 at 16:43
  • $\begingroup$ @DavidRaveh I am looking for a way to make sense of the author's justification. $\endgroup$
    – meth-
    Jan 8 at 16:47
  • $\begingroup$ Are you asking how the series being absolutely convergent allows him to rearrange the sequence? Sorry, it's a little unclear $\endgroup$
    – David Raveh
    Jan 8 at 16:49
  • $\begingroup$ I would say: compare with $\sum_{j=1}^\infty \frac{(-1)^n}n\stackrel?=\sum_{j=1}^\infty \frac{1}{2n}-\sum_{j=1}^\infty \frac1{2n+1}$. But the reality is that I won't be reading the book so chances are I will never know what rearrangements the author is referring to. $\endgroup$
    – Sassatelli Giulio
    Jan 8 at 17:01
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    $\begingroup$ If $\sum a_n$ and $\sum b_n$ are convergent then $\sum(a_n\pm b_n)$ are convergent and $\sum(a_n\pm b_n)=\sum a_n\pm \sum b_n.$ Apply this to $a_n={(iz)^n\over n!}$ and $b_n={(-iz)^n\over n!}.$ Then $e^{iz}+e^{-iz}=2\cos z$ and $e^{iz}-e^{-iz}=2i\sin z.$ From these two equations we get the formula $e^{iz}=\cos z+i\sin z.$ $\endgroup$
    – Ryszard Szwarc
    Jan 8 at 19:12

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Sorry. There is no "rearrangement" used here, it's just linearity of $\Sigma_0^\infty$.

Thanks for asking - I've been locked out, hadn't been checking every day, didn't realize I was back. yippee...

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  • $\begingroup$ Also, there are other instances where you write "rearrangement" in the textbook, so what do you exactly mean by rearrangements? $\endgroup$
    – meth-
    Jan 11 at 2:02

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