Rearrange Euler's identity to isolate i

Click For Summary
SUMMARY

The discussion centers on rearranging Euler's identity, specifically the equation e^{i \pi} + 1 = 0, to isolate the imaginary unit i. The transformation leads to the expression i = ln(-1)/π, which requires the use of complex logarithms due to the impossibility of taking the natural logarithm of a negative number in the real number system. Participants confirm that ln(-1) = iπ is valid within the context of complex analysis, emphasizing the importance of understanding complex logarithms for this manipulation.

PREREQUISITES
  • Understanding of Euler's identity and its implications in complex analysis
  • Familiarity with logarithmic functions, particularly complex logarithms
  • Knowledge of the properties of imaginary numbers and their representation
  • Basic grasp of mathematical notation and equations
NEXT STEPS
  • Study the properties of complex logarithms and their applications
  • Explore the implications of Euler's formula in various mathematical contexts
  • Learn about the branch cuts in complex analysis and their significance
  • Investigate the geometric interpretation of complex numbers on the Argand plane
USEFUL FOR

Mathematicians, physics students, and anyone interested in advanced mathematical concepts involving complex numbers and logarithms.

Gondur
Messages
25
Reaction score
0

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



e^{i \pi} + 1 = 0

The Attempt at a Solution



e^{i \pi} + 1 = 0

e^{i \pi} = -1

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

ln(e^{i \pi})=ln(-1)

i \pi=ln((-1))

i=\frac{ln(-1)}{\pi}
 
Physics news on Phys.org
You would have to take the complex logarithm, which is a subtle little thing.

https://www.physicsforums.com/showthread.php?t=637214
 
Last edited by a moderator:
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.
 

Similar threads

Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
18
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
6K
Replies
7
Views
2K
Replies
2
Views
2K