E to the pi * i

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  • #1
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[tex][/tex]

1. Compute all the values of [tex] e^ {\pi i} [/tex], indicating clearly whether there is just one or many of them.

Trivially, exp(pi * i) = -1. However, we can also consider e to be the complex number z, and pi * i to be the complex number alpha. Then we get:

[tex]e^{\pi i} = z^{\alpha} = e^{\alpha log(z)}
= e^{\alpha (Log |z| + i arg(z))}
= e^{\pi i (Log |e| + i arg(e))}
= e^{\pi i (1 + i2k\pi)}
= e^{\pi i}e^{-2\pi^{2}k}
= - e^{-2\pi^{2}k}
[/tex]
where k is an integer.

So what exactly is going on here? does exp(pi*i) = -1 or -exp(-2kpi^2)??

P.S. I hope all this tex doesn't mess up :(
 

Answers and Replies

  • #2
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something is wrong with LaTeX... it isn't displaying my tex right...
 
  • #3
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Fixed your LaTeX.
1. Compute all the values of [tex] e^ {\pi i} [/tex], indicating clearly whether there is just one or many of them.

Trivially, exp(pi * i) = -1. However, we can also consider e to be the complex number z, and pi * i to be the complex number alpha. Then we get:

[tex]e^{\pi i} = z^{\alpha} = e^{\alpha log(z)}
= e^{\alpha (Log |z| + i arg(z))}
= e^{\pi i (Log |e| + i arg(e))}
= e^{\pi i (1 + i2k\pi)}
= e^{\pi i}e^{-2\pi^{2}k}
= - e^{-2\pi^{2}k}
[/tex]
where k is an integer.

So what exactly is going on here? does exp(pi*i) = -1 or -exp(-2kpi^2)??

P.S. I hope all this tex doesn't mess up :(
 
  • #4
299
1
Is the complex exponential function invertible? (What is required for a function to have an inverse?)
 
  • #5
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Is the complex exponential function invertible? (What is required for a function to have an inverse?)
The function must be 1-1, right?
 
  • #6
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The function must be 1-1, right?
Correct. Does the complex exponential satisfy this?
 
  • #7
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Correct. Does the complex exponential satisfy this?
Sorry, I misread your first question. So, no, the complex exponential is not an invertible function. Where does my initial post break down then?
 
  • #8
299
1
Sorry, I misread your first question. So, no, the complex exponential is not an invertible function. Where does my initial post break down then?
When you tried to invert it.
 

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