Complex logarithms issue

Question

So, we know $\log(x)$ is real when $x>0$ and it has an imaginary component when $x<0$: \begin{equation}\log(x) =\log(e^{i\theta}|x|) = i\theta + \log(|x|) \end{equation}

However \begin{equation} \log(x) = \frac{1}{2} \log(x^{2}) \end{equation} Maybe there is an issue here (or else I don't know where it is), so point it out please.

In particular when $x$ is real: \begin{equation} \log(x) = \frac{1}{2} \log(x^{2}) = \frac{1}{2} \log(|x|^{2}) = \log(|x|) \end{equation} And that's quite weird.

In general: \begin{equation} \log(e^{i\theta}|x|) = \frac{1}{2} \log(e^{2i\theta}|x|^{2}) = i\theta + \frac{1}{2}\log(|x|^{2}) \end{equation}

However if $\theta=\{0,\pi\}$, then $e^{2i\theta}=1$, in the common case where we have a $2\pi$ redundancy, or how it's called.

So: \begin{equation} \log(-x) = i\pi + \log(|x|) = \log(|x|) \end{equation} \begin{equation} \log(x) = \log(|x|) \end{equation}

Where is the issue in these statements? I know there should be at least a big one, don't know where

Answer 1

\begin{equation}\log(x) =\log(e^{i\theta}|x|) = i\theta + \log(|x|) \end{equation}

The first mistake is here, since $e^{i\theta}=e^{i\theta+i2k\pi}$, then you get

\begin{equation}i\theta + \log(|x|)=i\theta+i2k\pi+\log(|x|) \end{equation}

which doesn't hold for $k\neq 0$

However \begin{equation} \log(x) = \frac{1}{2} \log(x^{2}) \end{equation}

The second mistake is here. Let $x=-1$

\begin{equation} \log(-1) = \frac{1}{2} \log((-1)^{2})=0 \end{equation}

which doesn't hold.