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Why doesn't e^i equal 1?

GROUPS:
Everyone knows that e^[pi(i)] == -1
Using this we also know that e^[2pi(i)] == 1
What 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 follows
i*[(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?
POSTED BY: Kaleb Burklow
Answer
10 months 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 BY: Shenghui Yang
Answer
10 months 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]= True

In[3]:= Simplify[(E^a)^b == E^(a b), Assumptions -> Element[b, Integers]]

Out[3]= True
POSTED BY: Ilian Gachevski
Answer
10 months ago