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Rearrange Euler's identity to isolate i

  • Thread starter Gondur
  • Start date
  • #1
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Homework Statement



Maybe this is not possible because i does not represent anything quantile and is merely abstract? I'm not sure and maybe you guys can help!


Homework Equations



[tex] e^{i \pi} + 1 = 0[/tex]

The Attempt at a Solution



[tex] e^{i \pi} + 1 = 0[/tex]

[tex] e^{i \pi} = -1[/tex]

You cannot take natural log of a negative number so where do I go from here?

[tex]ln(e^{i \pi})=ln(-1)[/tex]

[tex]i \pi=ln((-1))[/tex]

[tex]i=\frac{ln(-1)}{\pi}[/tex]
 

Answers and Replies

  • #2
22,097
3,282
You would have to take the complex logarithm, which is a subtle little thing.

https://www.physicsforums.com/showthread.php?t=637214 [Broken]
 
Last edited by a moderator:
  • #3
1,331
45
It is true that ln(-1) = ipi. I don't see anything wrong with what you've said.

You cannot take the ln of a negative in the reals. We are explicitly not limited to the reals.
 

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