Intuition for Euler's identity

  • Context: Undergrad 
  • Thread starter Thread starter Prem1998
  • Start date Start date
  • Tags Tags
    Identity Intuition
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
Prem1998
Messages
148
Reaction score
13
I read an intuitive approach on this website. You should read it, it's worth it:
https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/

I read that an imaginary exponent continuously rotates us perpendicularly, therefore, a circle is traced and we end up on -1 after rotating through pi radians.
If that's true, then why doesn't e^-pi rotates us continuously through 180 degrees so that we end up on the negative axis? '-' has more rotating power than 'i',right?
And, if that's not true, then please share your intuition of the formula.
 
Mathematics news on Phys.org
Prem1998 said:
I read an intuitive approach on this website. You should read it, it's worth it:
https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/

I read that an imaginary exponent continuously rotates us perpendicularly, therefore, a circle is traced and we end up on -1 after rotating through pi radians.
If that's true, then why doesn't e^-pi rotates us continuously through 180 degrees so that we end up on the negative axis? '-' has more rotating power than 'i',right?
No. The real part of the exponent does no rotation at all. Suppose we separate the exponent x+iy, into its real part, x, and its imaginary part, iy. Then ex+iy = exeiy, where the factor ex is the usual real exponential and eiy is a pure rotation in the complex plane. ex is just a pure multiplier with no rotation of the vector eiy in the complex plane. So e does no rotation at all.