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

[rustynail]

Hello,

I was playing around with DeMoivre's formula

e^i*pi = -1

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

e^2i*pi = 1

ln[e^2i*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[e^2i*pi]=0.

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

Thank you in advance for your time.

 

[WannabeNewton]

https://www.physicsforums.com/showthread.php?t=637214 It says micromass but it was really me...>.> ok it was micromass

 

[rustynail]

Great, I appreciate, thank you!