# Homework Help: (2pi)i=0? Confused about Complex Number

1. Mar 22, 2010

### mmmboh

I am confused about something, this isn't homework I was just fooling around with complex numbers, and found this:

$$e^{2\pi i}=1$$ so
$$ln e^{2\pi i}=ln 1=0= 2\pi i$$

Can someone explain this? the $$2\pi i=0$$ part...I must have done something illegal...

2. Mar 22, 2010

### Office_Shredder

Staff Emeritus
You're assuming that ln is a normal function. What's happening is similar to when you define arctan(x). tan(x) isn't invertible, but you pick a part of the graph that is invertible and take the inverse of that. Similiarly, exponentiation isn't actually invertible, and the reason why is because it's periodic: if you add $$2\pi i$$ to the exponent, you get the same thing again. So when you want to define ln, you have to restrict what values the imaginary part can take (the standard choice is that the imaginary part is between 0 and $$2\pi$$