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We know from Euler's formula that for $\theta \in \mathbb{R}$ we have $e^{i\theta} = \cos(\theta) + i \sin (\theta)$ which is a single complex number.

However, If I apply the definition of complex power $z^\alpha = e^{\alpha \log z}$ I get the following for $z=e$ and $\alpha = i \theta$:

$e^{i \theta} = e^{i \theta \log e} = e^{i \theta (Log|e| + i\arg e)} = e^{i\theta (1+2k\pi i)} = e^{-2k\pi \theta} \cdot e^{i\theta}$ with $k \in \mathbb{Z}$

This means that $(1- e^{-2k\pi \theta}) \cdot e^{i \theta} = 0$.

We know that $e^{i \theta}$ cannot be zero. So it must be that $1- e^{-2k\pi \theta} =0$ for all real $\theta$. This means that $k=0$. So $e^{i\theta} = e^{-2k\pi \theta} \cdot e^{i\theta}$ does not hold for all $k \in \mathbb{Z}$. What is going on here?

Also $e^{i \theta}$ is supposed to take infinitely many values since the exponent is neither a real integer, nor a rational fraction. Which way is it?

EDIT:

If I input $e^{\pi i}$ to Wolfram Alpha:

https://www.wolframalpha.com/input/?i=e%5E(pi*i)

I get both a single valued result -1 and a multivalued result.

fornit
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  • $e^{i \theta}$ is a single complex number. – copper.hat Jan 05 '18 at 19:58
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    If you write it as $\exp(i\theta)$ you are unlikely to suffer from such confusions. – Angina Seng Jan 05 '18 at 19:58
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    Your argument shows that the formula $z^\alpha = e^{\alpha\log(z)}$ is not always correct. – LSpice Jan 05 '18 at 20:02
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    There's a strong convention that $e^w$ always means $\exp(w)$, so unlike for other bases $e^w$ would have a single value also for non-integral $w$. But that convention ought to be stated before it's used. Euler's formula is unambiguously correct in the form $\exp(iz) = \cos z + i \sin z$ for all $z\in \mathbb{C}$. – Daniel Fischer Jan 05 '18 at 20:03
  • $e^{i\theta}$ is a perfectly well defined complex number. Things got messy only after you "applied the definition of complex power". Don't, please. –  Jan 05 '18 at 20:04
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    @ProfessorVector Why not? I am just applying definitions and as a result I get contradictions. – fornit Jan 05 '18 at 20:10
  • The comment of @LSpice says it all. – Lubin Jan 05 '18 at 20:59

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