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We know Euler's formula $$e^{i \theta} = \cos(\theta) + i \sin(\theta)$$

Let say $$\theta=\frac{\pi}{2}$$
Then we will get $$e^{i \frac{\pi}{2}} = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})$$
As we can see $$e^{i \frac{\pi}{2}} = i$$ Then we can wright... $$i^i = i^{e^{i \frac{\pi}{2}}}$$ or $$i^i = (e^{i \frac{\pi}{2}})^i = e^{-\frac{\pi}{2}} \approx 0.208$$
As we can see the result is as Real number. It is not an imaginary number!
Did I compute it correctly?
I think that I made little mistake above instead of $\theta=\frac{\pi}{2}$ I should wrote $\theta=\frac{\pi}{2} +2\pi k $ , $ k\in N\cup {0}$
$$e^{i {(\frac{\pi}{2}+2\pi k})} = i$$ $$i^i = (e^{i (\frac{\pi}{2}+2\pi k)})^i = e^{-\frac{\pi}{2} - 2\pi k}$$ So... $$k = 0 \implies e^{-\frac{\pi}{2}} \approx 0.208 $$ $$k = 1 \implies e^{-\frac{\pi}{2} - 2\pi} \approx 3.882*10^{-4} $$ $$k = 3 \implies e^{-\frac{\pi}{2} - 6\pi} \approx 1.354*10^{-9} $$

Is it true that: As we can see $ {i}^{i} $ has a lot of answers!?
I mean it has as many answers as many natural number exists!
Are all answers as a real number?

IremadzeArchil19910311
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  • It's understand that unlike real exponents imaginary powers have multiple values. I believe this is true regardless of base. I believe $2^i$ has infinite many values too. – fleablood Oct 03 '15 at 09:29
  • The trouble starts at "Then we can wright..." – Did Oct 03 '15 at 09:58

1 Answers1

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The mapping $z \mapsto z^\alpha$ is defined by $f_{\alpha}(z) = e^{\alpha\log(z)}$, note that the complex logarithm is defined with $\log(z) = \log |z| + i\arg(z)$

That being said, we can calculate: $i^i = e^{i\log(i)} = e^{i ( \log|i| + i\arg (i)) } = e^{-1\cdot(\frac{\pi}{2} + 2\pi k)} $.

  • There are infinitely many results. This is due to the multivalueness of the $\log , \arg$ functions.
  • As you can see, the result is real for all $k \in \mathbb Z$.
Ranc
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