does the identity $ln(z^n) = n*ln(z)$ hold for complex variables ?! This was a problem in a complex textbook:
is the collection of all values of $ln(i^2)$ the same as the collection of all values of $2ln(i)$?
And I guess the answer is no as the first will yield the set of values $i(\pi + 2k\pi)$ while the second yields $2ln(i) = 2(ln(1) + (\pi/2 + 2k\pi) i) = i(\pi + 4k\pi)$ which contradicts the identity given in Dennis Zill's textbook page 182 stating that
$(iii) lnz^n = nlnz$.

Just a random thought came to my mind after writing this question.
the first identity on the same page states that $(i) ln(z_1 z_2 ) = lnz_1 + ln z_2$. this and the third identity seem to be equivalent by substituting both $z_1$ and $z_2$ by $i$.
So, $$ln(i * i) = ln(i) + ln(i) = i*(\pi/2 + 2k\pi) + i * (\pi/2 + 2n\pi) = i * \pi + 2s\pi$$ which is correct but this implies one of two things:


*

*$ln(i) + ln(i) != 2ln(i)$

*$2*ln(i)$ is not a multiple of the multi-valued function $ln$ but rather a sum of two instances of $ln(i)$.
Which both seems very odd, So what am I getting wrong?!
Edit
n is an integer and ln is the multivalued function while Ln is the principal value
 A: 
Hint: In general the following holds for  the multivalued function $\arg z$ 
  \begin{align*}
\color{blue}{\arg(z^2)=\arg(z)+\arg(z)\ne 2\arg (z)}\tag{1}
\end{align*}
  See for instance this paper.

The argument $\arg (z)$ of a non-zero complex number $z\in\mathbb{C}$ is a multi-valued function and plays a key role in understanding the properties of the complex logarithm and power functions.

Any number $z\in\mathbb{C}$ can be written as
  \begin{align*}
z=|z|e^{i\arg (z)}
\end{align*}
  where $\arg(z)$ is a multi-valued function given by
  \begin{align*}
\arg(z)=\mathrm{Arg}(z) + 2k\pi\qquad\qquad k\in\mathbb{Z}
\end{align*}
  and $\mathrm{Arg}(z)\in(-\pi,\pi]$ is the principal value of $\arg(z)$.
We can consider $\arg(z)$ as the set of values
  \begin{align*}
\arg(z)=\{\theta,\theta+2\pi,\theta-2\pi,\theta+4\pi,\theta-4\pi,\ldots\}\tag{2}
\end{align*}
  with $\theta=\mathrm{Arg}(z)$. With the representation (2) in mind we consider the equation
  \begin{align*}
\arg (z_1)=\arg (z_2)
\end{align*}
  as equality of sets. We conclude
  \begin{align*}
\arg(z)+\arg(z)&=(\mathrm{Arg}(z)+2k\pi)+(\mathrm{Arg}(z)+2l\pi)\qquad\qquad &k,l\in\mathrm{Z}\\
&=\mathrm{Arg}(z)+\color{blue}{2} m\pi\qquad\qquad &m\in\mathrm{Z}\\
2\arg(z)&=2(\mathrm{Arg}(z)+2n\pi)=\mathrm{Arg}(z)+\color{blue}{4}n\pi
\qquad\qquad &n\in\mathrm{Z}\\
\end{align*}
  Since multiplication of complex numbers implies adding the arguments the statement (1) follows.

We are now ready to analyse $\ln(z^n)$.

We obtain for $n\in\mathbb{N},n>1$
  \begin{align*}
\ln(z^n)&=\underbrace{\ln(z)+\ln(z)+\cdots +\ln(z)}_n\\
&=\left(\ln|z|+i\mathrm{Arg}(z)+2k_1\pi i\right)+\left(\ln|z|+i\mathrm{Arg}(z)+2k_2\pi i\right)\\
&\qquad+\cdots+\left(\ln|z|+i\mathrm{Arg}(z)+2k_n\pi i\right)& k_1,\ldots,k_n\in\mathbb{Z}\\
&=n\ln|z|+ni\mathrm{Arg}(z)+2m\pi i& m\in\mathbb{Z}\\
n\ln(z)&=n\left(\ln|z|+i\mathrm{Arg}(z)+2k\pi i\right)\\
&=n\ln|z|+ni\mathrm{Arg}(z)+2\color{blue}{n}k\pi i&\qquad\qquad k\in\mathbb{Z}
\end{align*}
  We conclude the following are not set equalities:
  \begin{align*}
\color{blue}{\ln(z^n)=\underbrace{\ln(z)+\ln(z)+\cdots +\ln(z)}_n\ne n\ln (z)}
\end{align*}

A: You can think of a complex logarithm not as a multivalued function with values in $\Bbb C$, but as usual single-valued fnction, but with with values in $\Bbb C/2 \pi \sqrt{-1}$; image is still an abelian group, and so $ln(z^s) = s \, ln(z)$ for $s \in \Bbb C/2 \pi \sqrt{-1}$ makes sense (and true).
$exp$ and $ln$ can be though as single-valued and even bijective functions
$$exp: \Bbb C/2 \pi \sqrt{-1} \to \Bbb C^*,$$ $$ln: \Bbb C^* \to \Bbb C/2 \pi \sqrt{-1} $$ 
which are homomorphisms of corresponding groups — additive $\Bbb C/2 \pi \sqrt{-1}$ and multiplicative $\Bbb C^*$. (if you want to obtain "usual" $e^z$ and ln(z) from these, you should precompose $exp$ with factorization, and lift values after applying $ln$) But when you take preimage of a point along projection $\Bbb C \to \Bbb C/2 \pi \sqrt{-1}$, it's not a subgroup, but a coset, so multiplying by $s$ in $\Bbb C$ will get you wrong answer; you should first multiply in factor, and then lift values.
A: Let us use the following symbology
$$
\left\{ \matrix{
  {\rm Ln}(z) = \ln \left| z \right| + i\,{\rm Arg}\left( z \right) = \ln \left| z \right| + i\,\left( {\arg (z) + 2k\pi } \right) = \ln (z) + i2k\pi  \hfill \cr 
  \ln (z) = \ln \left| z \right| + i\arg (z) \hfill \cr}  \right.
$$
so that ${\rm Ln}(z)$ is the Multivalued function and $\ln(z)$ the principal branch.
Then it is always true that
$$
\eqalign{
  & {\rm Ln}\left( {z\,w} \right) = \ln \left| {\,z\,w\,} \right| + i\,{\rm Arg}(z\,w) =   \cr 
  &  = \ln \left| {\,z\,} \right| + \ln \left| {\,w\,} \right| + i\,{\rm Arg}(z) + i\,{\rm Arg}(w) =   \cr 
  &  = \ln \left| {\,z\,} \right| + \ln \left| {\,w\,} \right| + i\,{\rm arg}(z) + i\,{\rm arg}(w) + i\,2\,\left( {k + l} \right)\,\pi  =   \cr 
  &  = \ln \left( z \right) + \ln (w) + i\,2\,k\,\pi  =   \cr 
  &  = {\rm Ln}\left( z \right) + {\rm Ln}\left( w \right) \cr} 
$$
If you take note of the $k$ and $l$ that appear in the third line, you 
will immediately notice where you went astray, and how to deal with $z^n$,
i.e.:
$$ \bbox[lightyellow] {  
\eqalign{
  & {\rm Ln}\left( {z^{\,n} } \right)\quad \left| {\;0 \le n \in \mathbb Z} \right.\quad  = {\rm Ln}\left( {\prod\limits_{1 \le j\, \le \,\,n} z } \right) = n\,{\rm Ln}\left( z \right) =   \cr 
  &  = n\,\ln \left( z \right) + i\,\bigcup\limits_{1 \le j\, \le \,\,n} {\left\{ {2\,k_j \,\pi } \right\}}  = n\,\ln \left( z \right) + i\,2\,k\,\pi  \cr} 
}$$
Answer to your comments
The core lays on understanding the definition of ${\rm Arg}$ (in my notation).
When we write 
$$
{\rm Arg}(z) = \,{\rm arg}(z) + 2\,k\,\pi \quad \left| {\;k \in \mathbb Z} \right.
$$
we actually mean that to ${\rm arg}(z)$ we shall add  set of multiples of $2\pi$.
Then
$$
\eqalign{
  & {\rm Arg}(z\,w) = {\rm Arg}(z) + \,{\rm Arg}(w) =   \cr 
  & {\rm arg}(z) + 2\,k\,\pi  + {\rm arg}(w) + 2\,l\,\pi \quad \left| {\;k,l \in \mathbb Z} \right. \cr} 
$$
because we are not obliged to choose the same $k$ for $z$ and for $w$,
and since
$$
\left\{ {\left( {k + l} \right)\quad \left| {\;k,l \in \mathbb Z} \right.} \right\} = \left\{ {k\quad \left| {\;k \in \mathbb Z} \right.} \right\}
$$
we conclude that
$$
{\rm Arg}(z\,w) = {\rm arg}(z) + {\rm arg}(w) + 2\,k\,\pi \quad \left| {\;k \in Z} \right.
$$
That is equivalent to write
$$
z\,w = \left( {ze^{\,i2k\pi } } \right)\left( {we^{\,i2l\pi } } \right) = zwe^{\,i2k\pi } 
$$
Repeating the above we reach to $z^n$ ($n$ natural integer).
While instead, for $n=1/2$ for example, we get
$$
z^{\,1/2}  = \left( {ze^{\,i2k\pi } } \right)^{\,1/2}  = z^{\,1/2} e^{\,ik\pi }  = z^{\,1/2} e^{\,ik\pi } e^{\,i2l\pi }  = \sqrt z \,e^{\,ik\pi }  =  \pm \sqrt z \,e^{\,i2k\pi } 
$$
from which the ${\rm Arg}$ follows easily.
That means that the division by $2$ is applied to all the members of the set $ \left\{ {k\quad \left| {\;k \in \mathbb Z} \right.} \right\}$,
differently from the multiplication by $n$, where instead the set is composed by adding $n$ different elements.
I do know that the subject is not easy. I do not know the sources that you mention. Mine studies on the subtleties of exponentiation and log 
were on " Theory of analytic functions" - A.I. Markuševič
