I Am I misapplying something here? (Exponentials; Euler's identity)

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Euler's identity states that exp(iπ) = -1, leading to exp(2iπ) = 1. The discussion highlights a potential misunderstanding when applying natural logarithms, suggesting that ln(exp(2iπ)) = 0, which implies 2iπ = 0. This confusion arises from the properties of logarithms and exponentials, particularly in the complex plane. The inverse of the exponential function is multi-valued, meaning that if e^x = 1, x can equal n2πi for any integer n. The key takeaway is that the complexities of complex numbers can lead to seemingly paradoxical conclusions if not carefully considered.
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Reconciling an apparent (?) paradox in an exponential
We all know from Euler's identity that exp(iπ)=-1. And from the laws of exponents, exp(2iπ)=(exp(iπ))²=1.

Further, for any real number a≠0, a⁰=1.

Then, since two things equal to the same third thing are necessarily equal, exp(2iπ)=(exp(iπ))²=a⁰=1.

Here's where I'm wondering if I've stripped one or more intellectual gears. If we now take natural logarithms, it would seem we get

ln(exp(2iπ)) = 2iπ = 0*ln (a) = 0

That would suggest at first glance that 2iπ = 0, which made me do a double take, to say the least. I realize this may well be fallacious, and I may be guilty of missing something obvious and fundamental. Help me out here (gently, if you would, please): what did I miss?
 
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In general, ##f^{-1}(f(z))## is not necessarily equal to ##z##. Can you see why?
 
Without using complex numbers at all, you could have said that ##(1)^2 = 1 = 1^0##, so ##2=0##.
The truth is that the inverse of the exponential function is best understood as a multi-valued function in the complex plane. If ##e^x = f(x) = 1##, then ##x## can equal ## n2\pi i ##, for any integer, ##n##.
 
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