How to solve $\ln x^n=0$?
This is because you are not specifying where you want to solve your equation.
Since $\ln$ is usually defined on $\mathbb {R}_+^*$, you can not accept solutions outside of that set, in your case: $\mathbb C$.
But if you give yourself a definition of the logarithm on $\mathbb C$, then you can also find your complex solutions.
To answer a comment:
If you take this as a definition for a complex logarithm:
$$\log(z)=\ln(z)+i\arg(z),$$
then you get
$$\begin{align*} \log(z^n)=0&\iff \ln\vert z^n\vert+i\arg(z^n)=0 \\ &\iff n\ln\vert z\vert+ni\arg(z)=0\\ &\iff \begin{cases} \vert z\vert =1 \\ n\arg(z)=0 \pmod {2\pi} \end{cases} \\ &\iff \begin{cases} \vert z\vert =1 \\ \arg(z)=0 \pmod {\frac{2\pi}n}, \end{cases}\end{align*}$$
which gives you the solution you want.
One should be sure what definition one is using for the symbol "$\ln$".
- In real analysis, the single-valued real logarithm $\ln(x)$ is only defined for $x>0$;
- In complex analysis, $\ln(x)$ is usually reserved for positive real number $x$ while the multi-valued complex logarithm is denoted as $\log(z)$ for non-zero complex number $z$. The relation between "$\ln$" and "$\log$" is then given by the definition $$ \log(z)=\ln(|z|)+i\arg (z),\quad z\in{\bf C}\backslash\{0\}\tag{*} $$ where $ \mathrm{arg}(z) := \{ \theta \in {\bf R}: \cos \theta + i \sin \theta = \frac{z}{|z|} \} $ denotes all the possible arguments of ${z}$ in polar form.
- Of course you could see that some people insists on using $\ln(z)$ for complex number $z$ as well, which would cause unnecessary confusion regarding what the value of $\ln(z)$ should be when $z$ is a positive real number.
Now, let's stick to the definition and also the symbols in (*), and see what should be the solution to $\log (z^n)=0$.
By definition, $$\ln(|z^n|)+i\arg(z^n)=0$$ which implies that $|z^n|=1$ and $\arg(z^n)=0$. It means that $z^n=1$ and now you can use the De Moivre's formula. This is essentially what you did in Method 1.
The problem in Method 2 is that the identity $\ln a^b=b\ln a$ $(a,b>0)$ is only true for the real logarithm, which would a false assumption if you are in the complex world. For instance, it is an instructive exercise to check by definition of the complex logarithm that $$ \log (-1)^2\neq 2\log(-1). $$