From Euler's identity: i^i=exp(-pi/2)= 0.2079 (rounded)

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The discussion centers on Euler's identity, specifically the expression i^i = exp(-π/2) ≈ 0.2079. Participants clarify that i^i does not have a unique value due to the multi-valued nature of the complex logarithm, represented as i^i = e^{i^2(π/2 + 2πn)} for all integers n. The conversation emphasizes the importance of using LaTeX for mathematical notation to enhance clarity in discussions.

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SW VandeCarr
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From Euler's identity: i^i=exp(-pi/2)= 0.2079 (rounded). I've always thought of this as an interesting result although I don't know of any particular significance or consequence of it. Is there any?
 
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i^i does not have a specific value. i^i=e^{i^2(\frac{\pi}{2}+2\pi \cdot n) }=e^{-(\frac{\pi}{2}+2\pi \cdot n) } for all integers n.
 
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Jarle said:
i^i does not have a specific value. i^2=e^{i^2(\frac{\pi}{2}+2\pi \cdot n) }=e^{-(\frac{\pi}{2}+2\pi \cdot n) } for all integers n.

I'm using ^ as raising to a power except for 'exp' where exp(x) means e^x

exp(i pi)= -1

SQRT [exp(i pi) = SQRT (-1)

exp(i pi/2) = i

exp ((i^2) pi/2)) = i^i = exp (-pi/2) = 0.2079 (rounded)
 
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SW VandeCarr:

Jarle is also using it in that sense.

However, as he pointed out, the complex logarithm is a multi-valued mapping, in contrast to the real logarithm.
 


arildno said:
SW VandeCarr:

Jarle is also using it in that sense.

However, as he pointed out, the complex logarithm is a multi-valued mapping, in contrast to the real logarithm.

Thanks, but the algebra is correct, is it not? Normally I don't see the +2pi.n term in texts.
 
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Strange, I do!
 


SW VandeCarr - you can put "tex" tags around your equations and use latex syntax instead of defining all of your notation. It makes yours and everyone else's life easier :)
 


daviddoria said:
SW VandeCarr - you can put "tex" tags around your equations and use latex syntax instead of defining all of your notation. It makes yours and everyone else's life easier :)

Thanks daviddoria. I guess it's about time I started using latex if I'm going to be posting questions here.
 

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