e^(i*pi) = -1 paradox?

Question

I have 2 paradoxes involving the formula e^(i*pi) = -1
-1 = e^(i*pi) taking the natural log of both sides gives
ln(-1) =i *pi. Now for the paradox
0 =ln(1) = ln( (-1^2) ) =2*ln(-1) = 2*i*pi
So that means 0 =2*i*pi
Dividing both sides by 2*pi gives
0 =i


Second Paradox
e^(i*pi) = -1 Take the reciprocal of both sides gives
e^(-i*pi = -1
Therefore, e^(i*pi) = e(-i*pi)
Take the natural log of both sides gives
i*pi = -i*pi
Dividing by i*pi gives
1 =-1

Answer 1

The problem is not that you can't take the log of a complex number. The problem is that log is multiple valued in the complex plane. In other words, the exponential function is not one-to-one. You can take the 'principle branch' of the logarithm, but then you can lose certain of the rules of logarithms.

The upshot is that when you take the log of -1, you either get an infinite set of values (.., -pi*i, pi*i, ...) and another infinite set of possibilities for the log of 1 OR you lose the property that log(a^b)=b*log(a) that you used in your first paradox. This is just a fact of life when working with complex numbers. In the second, BOTH -pi*i and pi*i are logarithms of -1, so the argument fails. It's somewhat like saying that the squares of -1 and 1 are the same, so -1=1.

Answer 2

i*pi is a complex number,
so you cannot write ln(-1) = i*pi,
because the natural logarithm lnx is defined only for x>0.
If you want to write the equation using the complex logarithm, then you must specify which branch you are using.

See the following discussion:
http://mathworld.wolfram.com/PrincipalValue.html

The following will show how standard branches (Principal banches) of multi-valued functions are defined using branch cuts:

http://mathworld.wolfram.com/BranchCut.html

Answer 3

ok, no, your looking at eulers formula, and its basically, you have to take into account the sine and cosine forms for imaginary numbers, then you get -1 as a result, a search on google, or a better site, mathworld.com will show you.

the second one, same thing, use the sine and cosine forms for imaginary numbers, and it works

hope this helps, matttlocke

Answer 4

You cannot take the log of a negative number.