Unraveling Euler's Formula: e^(pi*i) +1 = 0

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Discussion Overview

The discussion centers around Euler's formula, specifically the equation e^(pi*i) + 1 = 0. Participants explore its implications, representations, and related mathematical concepts, including complex exponentiation and logarithms.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant shares their interpretation of Euler's formula and attempts to express it in a different format, questioning its correctness.
  • Another participant confirms the standard form of the equation and notes that it has been discussed previously on the forum.
  • A participant inquires about the familiarity of their specific representation involving roots and complex numbers.
  • Another participant highlights that e^(pi) is not the only solution for (-1)^(1/i) and suggests exploring other values that arise from different branches of the logarithm.

Areas of Agreement / Disagreement

Participants generally agree on the validity of Euler's formula, but there are differing views on the specific representation and implications of complex exponentiation. The discussion remains unresolved regarding the exploration of other values related to the logarithm.

Contextual Notes

There are limitations regarding the assumptions made about the branches of logarithms and the interpretations of complex exponentiation, which are not fully explored in the discussion.

strid
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Just wanted to share a cool thing I found when I was shown Eulers formula...

e^(pi*i) +1 = 0

this can be written as

e^pi = (SQRT -1)ROOT -1

dont know if I wrote correclty...

.................
.................
...#######...########################......
.##...#...#...#...............
...#.#...--.#...#......#.........
...#...#... #......#.........
......#...----...#..........
..########...#......#............
....#...#......#.........
.....#.#.....#.........
......#.............
.................

Took a time to write this :)


so can you imagine the number above to equal e^pi which is th real number around 23...? :smile:
 
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Yes,
[tex]e^{i\pi}+ 1= 0[/tex]

(click on that to see the code I used)

I suspect that just about everyone on this board already knows that- there have been a number of threads about it.
 
but is the format i wrote it in also well-known? the cool root stuff I mean...
 
Yes it is. Exponentiation by complex numbers gives some startling results when you first see it.

Note that [tex]e^{\pi}[/tex] isn't the only answer for [tex](-1)^{1/i}[/tex], it depends on the branch of the logarithm you used. Can you find the other values? They might be even more suprising...
 

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