E to the pi i

  • #1
ok, here it goes, why is e^(pi.i)=-1 ?
 
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Answers and Replies

  • #2
Integral
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It comes form Eulers relationship

[tex] e^{ix} = \cos(x) + i \sin(x) [/tex]

Edit: LOL, I fixed it already!
Now I am moving this to Math.
 
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  • #3
freemind
Any complex number [tex] a + bi [/tex] can be written in the form [tex] r(\cos{\theta} + i\sin{\theta}) [/tex]. De Moivre's theorem states that this in turn can be written in the form [tex] re^{i\theta} [/tex]. That is, [tex] re^{i\theta} = r(\cos{\theta} + i\sin{\theta}) [/tex]. Plug in [tex] \theta = \pi [/tex] and behold the magic!
 
  • #4
Was it Euler? well, in that case u would have to use Taylor's expansion series and yes, it would work. Why is this important at all then?
 
  • #5
freemind
It is commonly regarded to be one of the most (ahem) beautiful and elegant mathematical relationships in our universe (yes, our :biggrin: ). C'mon, wouldn't you agree that it is beautiful, succintly relating the 5 most important numbers in mathematics?!?! ([tex] e^{\pi i} + 1 = 0 [/tex])
 
  • #6
Integral
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freemind said:
It is commonly regarded to be one of the most (ahem) beautiful and elegant mathematical relationships in our universe (yes, our :biggrin: ). C'mon, wouldn't you agree that it is beautiful, succinctly relating the 5 most important numbers in mathematics?!?! ([tex] e^{\pi i} + 1 = 0 [/tex])
In addition it uses each of the fundamental mathematical operations, addition, multiplication, exponentiation, and equality. It is consider mathematical poetry.
 

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