Power Rule for Complex Logarithms

Question

Given $z \in \mathbb{C}$, could you explain to me why: $$ \log(z^n) \neq n\log(z) $$ in terms of analytic branches of $\log$ functions?

Answer 1

Let us consider the principal branch of the logarithm, namely \begin{align} \operatorname{Log} z= \log|z|+i\arg z \ \ \text{ where } -\pi<\arg z<\pi \end{align} for all $z \in \mathbb{C}\backslash\{(-\infty, 0]\}$.

Clearly, we see that \begin{align} \operatorname{Log} (-1)^2 =\operatorname{Log} 1= \log 1 + i\arg 1 = 0 \end{align} but $2\operatorname{Log}(-1)$ doesn't make sense.

Additional Example: Let $z = i$, then $z^4 = 1$. Next, observe \begin{align} \operatorname{Log} 1 = 0 \neq 4\operatorname{Log} i = 4\left( \log|i|+i\arg i\right) = 4i\frac{\pi}{2} = 2\pi i. \end{align}

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\ln\pars{z} = \ln\pars{\verts{z}} + \arg\pars{z}\ic\,,\quad z \not= 0\,,\quad -\pi < \arg\pars{z} < \pi;\quad}$ see this example:

\begin{align} \ln\pars{-1 + \ic} + \ln\pars{-1 + \ic} & = \bracks{{1 \over 2}\ln\pars{2} + {3\pi \over 4}\,\ic} + \bracks{{1 \over 2}\ln\pars{2} + {3\pi \over 4}\,\ic} = \ln\pars{2} + \color{red}{{3\pi \over 2}\,\ic} \\[5mm] \ln\pars{\bracks{-1 + \ic}\bracks{-1 + \ic}} & = \ln\pars{-2\ic} = \ln\pars{2} \color{red}{- {\pi \over 2}\,\ic} \\[5mm] \implies & \bbx{\ln\pars{\bracks{-1 + \ic}\bracks{-1 + \ic}} \not= \ln\pars{-1 + \ic} + \ln\pars{-1 + \ic}} \end{align}