I am confused about the differences of the properties of natural log in complex analysis and in real region. This question might be a bit stupid, but any answers or explanations of the log properties will be appreciated.
The question is:
$\log(e^{2\pi i})=\log|e^{2\pi i}|+i \arg(e^{2\pi i})=\log|\cos(2\pi)+i \sin(2\pi)|+i2\pi=\log|1|+2\pi i=2\pi i$
$\log(e^{2\pi i})=2\pi i$ (since $\log(e^k)=k$)
$\log(e^{2\pi i})=\log(\cos(2\pi)+i \sin(2\pi))=\log(1+0i)=\log(1)=0$
Which ones of the three equations are correct?
And
$e^{2\pi i}=\cos(2\pi)+i \sin(2\pi)=\cos(0)+i \sin(0)=e^{0i}$, then $2\pi=0$
which step is wrong?
Thanks for your help!
Which one is correct depends on whether you are interested in the logarithm as a mapping from reals to reals, or from complexes to complexes.
If you want to think of the logarithm as a real-valued function, then you would write $e^{2\pi i} = 1$, then $\log (e^{2\pi i}) = \log 1 = 0.$ But if you want to think of it as a multi-valued mapping from $\mathbb C$ to $\mathbb C$, then $$\log z = \log |z| + i \arg(z) + 2 \pi i k, \quad k \in \mathbb Z,$$ and $$\log (e^{2\pi i}) = 2\pi i k$$ is a set of complex numbers, since the complex logarithm is a one-to-many mapping. However, for the purposes of calculation, we are often interested in choosing a branch of this mapping; e.g., we might choose the (unique) branch such that $$\Im(\log (z)) \in (-\pi, \pi]$$ for all $z \in \mathbb C$. In this case, you'd get $\log(e^{2\pi i}) = 1$, since the only choice for the argument that lies in $(-\pi, \pi]$ is $0$. But there is nothing intrinsically special about this choice.
The second part of your question suggests to me that you have a misconception. Here's a simple example. Let $$f : \mathbb R \to \mathbb R, \quad f(x) = x^2.$$ Now, we note for example, $$f(-2) = 4, \quad f(2) = 4.$$ But just because $f(-2) = f(2)$ does not mean that $-2 = 2$! So why would you conclude that $2\pi = 0$ based on your above reasoning?
The fallacy, as applied to $f(x) = x^2$, is that when we look at the inverse function $f^{-1}(x) = \sqrt{x}$, we frequently think of this as "take the nonnegative square root." But there is nothing special about this choice: $-2$ is as much a square root of $4$ in the sense that it is a legitimate solution to $x^2 = 4$ as $2^2 = 4$. In the complex numbers, this "multivaluedness" is even more apparent: note that $(1-i)^2 = 1^2 - 2i + (-i)^2 = -2i$, but we also have $(-1+i)^2 = (-1)^2 - 2i + i^2 = -2i.$ So if I ask you for "the square root" of $-2i$, which one would you choose? There is no such thing as the "nonnegative" root, since $\mathbb C$ is not an ordered field.
So, what if you say, "choose the root that has nonnegative real part." Okay, then under this rule, which of the five roots of $3+17i$ would you choose? By the pigeonhole principle, you are guaranteed to have at least two distinct complex roots with nonnegative real part.
This brings us back to the discussion of branches. The complex mapping $$g : \mathbb C \to \mathbb C, \quad g(z) = z^{1/n}$$ is a "one-to-n" mapping: for any nonzero $z$, there are $n$ images of $z$ under $g$. When we speak of taking the "nonnegative square root" for the case $n = 2$, we are actually choosing a branch. And when we do this, we are no longer allowed to make statements like $-2 = 2$ on the basis that $(-2)^2 = 2^2$. For the exact same reason, you cannot say $2\pi = 0$ simply because $e^{2\pi i} = e^{0i} = 1$.