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Why doesn't e^i equal 1?

Posted 11 years ago
 Everyone knows that e^[pi(i)] == -1Using this we also know that e^[2pi(i)] == 1What I believe I proved was e^i == 1 using the following proof:Start with [e^i]Since raising a number to 1 just gives you the number let's raise this to the equivalent of 1 or [(2pi)/(2pi)][e^i]^[(2pi)/(2pi)]When raising a power to a power you just multiply them to get the new power also when multiplying 2 number you can also pull them apart as followsi*[(2pi)/(2pi)] == [2pi(i)]*[1/2pi] returning to the original problem we now get [e^(2pi(i))]^[1/(2pi)] but since we already know that [e^(2pi(i))] == 1 we get the following[1^(1/{2pi})] but 1 raised to any power equal to 1 therefore we have e^1 eqaul to 1 which we know isn't correct.Where did I mess up?
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Posted 11 years ago
 Basically it is same question as: what is (-1)^(2/3). z^(1/r) is not defined usually with r being not integer.
Posted 11 years ago
 When raising a power to a power you just multiply them to get the new power As your example shows, this is not true for all complex powers. The statement is valid, however, if the inner power is a real number or the outer power is an integer: In[2]:= Simplify[(E^a)^b == E^(a b), Assumptions -> Element[a, Reals]]Out[2]= TrueIn[3]:= Simplify[(E^a)^b == E^(a b), Assumptions -> Element[b, Integers]]Out[3]= True