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Euler's formula states that: $$e^{iθ} = cos(θ) + i*sin(θ)$$ Plugging in 2π for theta gives 1, but so does plugging in 0 for theta. Therefore: $$e^0 = e^{2iπ}$$ Taking the natural logarithm of both sides gives: $$ 0 = 2πi$$ Dividing both sides by 2pi yields: $$0 = i$$.

Clearly, I have a mistake somewhere, but I don't know where. Am I misusing Euler's formula or is it something else? Thanks in advance for any help.

Darcy Sutton
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    Don't worry, you just need more experience working with complex numbers. Your problem is the step "Taking the natural logarithm of both sides". – GEdgar Oct 27 '22 at 20:59
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    BTW, $1^2 = (-1)^2$, but $1 \ne -1$ exhibits a similar phenomenon. – JonathanZ Oct 27 '22 at 21:06
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    Log in complex analysis is very weird and includes a branch (i.e. it’s discontinuous). I think the issue happened when you took the log of exp(i 2 pi). – NicNic8 Oct 27 '22 at 21:09

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Well problem is if $f(B)=f(A)$ ($f$ is a function), that does not always mean that $B=A$. For example $\sin(0)=\sin(2\pi)$; that doesn't mean that $0=2\pi$. In real numbers when $e^A=e^B$ ($A$ and $B$ are real numbers), we can say that $A=B$, but that is not always true for complex numbers.

bobeyt6
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Domahidi Péter
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