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If I write a real number, say $2$ in polar coordinate as $2e^{2k\pi\text{i}}$, where $k \in \mathbb{Z}$; and raise this form to $\text{i}$-th power and I get: $$ 2^\text{i}=\left(2e^{2k\pi\text{i}}\right)^\text{i}=2^\text{i} \cdot e^{-2k\pi} $$ Then, it seems that $k$ must be zero. Does it means I cannot write real number in polar form with $k=\pm1, \pm2, \cdots$? Or am I missing something. Is the statement $2=2e^{2\pi\text{i}}$ true then?

bostonyyang
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    Are you assuming that $\left(e^{2k\pi i}\right)^i=e^{-2k\pi}$? – José Carlos Santos Sep 17 '24 at 22:08
  • By definition

    $$z^w=e^{w\log z}\implies 2^i=e^{i\log 2}=\cos(\log 2)+i\sin(\log 2)$$

    but $\left( e^{2k\pi i}\right)^i$ is not well defined. Then yes is like to assume $k=0$.

     – user Sep 17 '24 at 22:26
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    What it means is $(a^b)^c=a^{bc}$ doesn't hold for complex exponentiation, something which has been discussed many times before on this website. – Gerry Myerson Sep 17 '24 at 23:48
  • Related: https://math.stackexchange.com/questions/2380281/the-proper-way-to-interpret-1m-n and https://math.stackexchange.com/questions/3343833/proof-that-sqrtx2-x-by-using-index-laws-why-is-this-wrong and https://math.stackexchange.com/questions/1755556/what-is-solution-of-j3-j-is-complex-number and https://math.stackexchange.com/questions/927054/why-is-eulers-formula-valid-for-all-n-but-not-de-moivres-formula and https://math.stackexchange.com/questions/3817464/question-about-the-fundamental-theorem-of-algebra and probably lots of others. – Gerry Myerson Sep 17 '24 at 23:53
  • Care to engage with us, boston? – Gerry Myerson Sep 20 '24 at 03:19
  • @JoséCarlosSantos: Thank you for pointing out I was missing the fact that $z^\text{i}$ has infinite branches. – bostonyyang Sep 21 '24 at 13:15
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    @GerryMyerson: Thank you for all the useful links that pointed me at the right direction. – bostonyyang Sep 21 '24 at 13:16
  • To summarize my understanding, and in general, for arbitrary choices of the pair $p,q\in\mathbb{Z}$, the statement $Ae^{2p\pi\text{i}}=Ae^{2q\pi\text{i}}$ is always true, because there always exists branches on both sides to satisfy the equality. – bostonyyang Sep 21 '24 at 13:22
  • If $p$ is an integer, then $e^{2p\pi i}=1$, and "branches" have nothing to do with it. – Gerry Myerson Sep 21 '24 at 14:33

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