Taking the natural logarithm of e^(2i*pi)

  • Thread starter rustynail
  • Start date
Hello,

I was playing around with DeMoivre's formula

ei*pi = -1

and there is something I don't quite understand about taking the natural logarithm of a certain expression. I though that

e2i*pi = 1

ln[e2i*pi] = ln (1),

but this yields to an imposibility

2i*pi = 0.

So obviously I am doing something wrong, and when I input ln[e^(2i*pi)] into Wolfram, it gives log[e2i*pi]=0.

Can anyone explain why Wolfram Alpha translates ln[e^(2i*pi)] to log[e2i*pi]=0, and why that second expression is true?

Thank you in advance for your time.
 

WannabeNewton

Science Advisor
5,774
527
https://www.physicsforums.com/showthread.php?t=637214 [Broken] It says micromass but it was really me...>.> ok it was micromass
 
Last edited by a moderator:
Great, I appreciate, thank you!
 

Related Threads for: Taking the natural logarithm of e^(2i*pi)

  • Posted
Replies
8
Views
2K
  • Posted
Replies
5
Views
507
Replies
16
Views
3K
P
  • Posted
Replies
14
Views
3K
  • Posted
Replies
6
Views
608

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top