does the identity $\ln(z^n) = n\times\ln(z)$ hold for complex variables ?!

Question

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 $2\ln(i)$?

And I guess the answer is no as the first will yield the set of values $i(\pi + 2k\pi)$ while the second yields $2\ln(i) = 2(\ln(1) + (\pi/2 + 2k\pi) i) = i(\pi + 4k\pi)$ which contradicts the identity given in Dennis Zill's textbook page 182 stating that

$(iii) \ln z^n = n\ln z$.

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Just a random thought came to my mind after writing this question.

the first identity on the same page states that $(i) \ln(z_1 z_2 ) = \ln z_1 + \ln z_2$. this and the third identity seem to be equivalent by substituting both $z_1$ and $z_2$ by $i$. So, $$\ln(i \times i) = \ln(i) + \ln(i) = i\times(\pi/2 + 2k\pi) + i \times (\pi/2 + 2n\pi) = i \times \pi + 2s\pi$$ which is correct but this implies one of two things:

1. $\ln(i) + \ln(i) \neq 2\ln(i)$

2. $2\times \ln(i)$ 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

Answer 1

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 $z\in\mathbb{C}\setminus\{0\}$ is a multi-valued function and plays a key role in understanding the properties of the complex logarithm and power functions.

Any number $z\in\mathbb{C}\setminus\{0\}$ 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*}

Answer 2

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.

Answer 3

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