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What is the equivalent to " $(e i)^z$ " , where i is the imaginary "i" and z is a variable (maybe a complex one) ? (I'm thinking in a possible symmetry with $e^{iz}$)

Luis
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  • Do you perhaps want to mean $(e^i)^z$ instead of $(ei)^z$? – Berci May 04 '14 at 21:54
  • No. I want to mean $(ei)^{z}$ not $e^{iz}$. – Luis May 04 '14 at 21:58
  • Why not directly $i^z$? – vonbrand May 05 '14 at 02:37
  • $i^x$ seems interesting to x $\in$ $\mathbb{R}$, I think that a curve generated is a circumference. But to x $\in$ $\mathbb{I}$ ? What curve result of this ? – Luis May 06 '14 at 22:09
  • Note that f(x) -the image - is plotted in a $\mathbb{R}$ $x$ $\mathbb{I}$ plane. – Luis May 06 '14 at 22:26
  • I think that $(ie)^{x}$ generates a circumference as $e^x$ to x $\in$ $\mathbb{I}$. The difference between the two curves is only the "phase" - I guess. – Luis May 06 '14 at 22:36
  • If I multiply a real by "i", I get a number in the $\mathbb{I}$ set. But if I multiply "i" by a real number, I get a number in the same set $\mathbb{I}$. If I multiply two numbers in the $\mathbb{I}$ set, a get another number in the $\mathbb{R}$ set, the same does not occur between two numbers in the $\mathbb{R}$ set. So, the symmetry between the $\mathbb{R}$ set and $\mathbb{I}$ set does not exist. – Luis May 06 '14 at 22:44

1 Answers1

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Here is how you advance

$$(ie)^z = e^{z\ln(ei)}=e^{z(\ln(e)+i(\frac{\pi}{2}+2k\pi i) } = e^{z(1 + \frac{i\pi}{2} - 2k\pi ) } = \dots,\quad k \in \mathbb{Z}. $$

Mhenni Benghorbal
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  • $$(ie)^z = e^{z\ln(ei)}=e^{z(\ln(e)+i(\frac{\pi}{2}+2k\pi i) } = e^{z(1 + \frac{i\pi}{2} - 2k\pi ) } = e^z . e^\frac{i\pi}{2} . e^{-\frac{2k}{\pi}} \quad k \in \mathbb{Z}. $$

     – Luis May 04 '14 at 22:22