What is the value of $\ln(e^{2i\pi})$? According to Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta;$ so, $e^{2\pi i} = 1.$ And $\ln(1) = 0.$
On the other hand, $\ln(e^{i\theta}) = i\theta,$ so $\ln(e^{2\pi i}) = 2\pi i$.
Is $\ln(e^{2i\pi})$ equal to $0$ or $2\pi i$?
 A: The natural logarithm is multivalued with respect to complex numbers. This means that the natural logarithm is not a function because its outputs are not unique. In order to deal with this, mathematicians tend to view what they call the "principal branch" of the natural logarithm or the main branch, denoted "$\operatorname{Ln} x$". While $\ln x$ is not a function, $\operatorname{Ln} x$ is. Therefore $\ln 1 = 0, 2\pi i,\dots$. In fact, it is equal to $2\pi in$ where $n$ is an integer all at the same time. It's not just 1 that has this property; all complex values for which the natural logarithm is defined share this property. However, given how it is generally defined, $\operatorname{Ln} x = 0$ only. If you know about $\arcsin$ and $\arccos$, it's kind of like that.
Great question. Let me know if I wasn't clear or if I got something wrong.
Also, as a side note, when using the $ for the math editing tool, you can write "\ln" in order to make it straight.
A: Please consider the fact that
\begin{equation}
\log z = \log |z|+j (arg(z)+2n \pi)
\end{equation}
where $z \in \mathbb{C}$, and $n=0,\pm 1,\pm 2, ...$, and $j=\sqrt{-1}$. As you wrote $|e^{2 \pi j}|=1$, while $arg(e^{2 \pi j})=0$. The complex argument can be computed by using the arc tangent function of the ratio between imaginary part and real part of the complex number. When doing so we can conclude that
\begin{equation}
\log e^{2 \pi j} = 2n \pi j
\end{equation}
When n = 0  we refer to as the principal branch of the
logarithm; the corresponding value of the function is referred to as the principal value. If z is a real and positive number, we usually take $n = 0$ so that the principal branch of the complex logarithm function agrees with the usual one for real variables. In reality
the function $\log z$ is infinitely branched.
