Another fallacy

1. Jun 14, 2005

mathelord

HI, since e^[i*pi]=-1,then e^[2i*pi]=1.
taking lin of both sides,2*i*pi=0,that makes i=0
squaring both sides,you get -1 to equal 0.
Is there a problem with what i have done,and what is it?

2. Jun 14, 2005

Curious3141

When complex outputs are allowed, logarithms become multivalued. In this case, the correct formulae are :

$$e^{2k\pi i} = 1, \ln(1) = 2k\pi i$$ where $k$ ranges over the integers.

In your case, you're confusing the result for k = 0 with k = 1, hence the apparent paradox.

Last edited: Jun 14, 2005
3. Jun 14, 2005

Zurtex

Yes and:

$$(-1)^2 = 1 \quad 1^2 = 1 \quad 1 = - 1 \quad 2 = 0$$

It's the same things really