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

#### rustynail

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?

#### WannabeNewton

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:

#### rustynail

Great, I appreciate, thank you!

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