Using log on Euler's formula to get i=0?

Question

By Euler's identity,

$ e^{2πi} = 1 $

$ 2πi = log_e{1} $

$ i = 0 $

What am I doing wrong? Is any domain being violated? What does this actually mean?

Answer 1

Notice that

$$e^{i2k\pi}=1$$ for any integer $k$ (which you wrote for $k=1$). Then taking the logarithm function,

$$\log e^{i2k\pi}=\log1$$ must hold. This means that if you accept that

$$\log e^{i2k\pi}=i2k\pi$$ then perforce

$$\log 1=i2k\pi.$$

This is not paradoxical. As the logarithmic equation has infinitely many solution, you have to select some "branch" to define a function. You are free to pick the branch such that

$$\log 1=i2k\pi,$$ (where $k$ is chosen once for all) provided you stick to it.

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Also notice that for this reason, the rule

$$\log ab=\log a+\log b$$ is lost in the complex.

Answer 2

From the equality $e^{2\pi i}=1$, what you can deduce is that $2\pi i$ is a logarithm of $1$. It turns out that $0$ is another logarihtm of $1$. That's all. Every non-zero complex number has infinitely many logarithms.