Why Does \( e^{2\pi i} = 1 \) Lead to Confusion About \( 2\pi i = 0 \)?

mmmboh
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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...
 
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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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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