WolframAlpha seems to suggest that $log(z) = log(-z)$

Question

Wolfram-Alpha returns this snippet at the end of the "step by step solution":

I'm confused by the negation of the expression inside the logarithm. The definition of complex logarithm is:

$$log(z) = ln|z| + i (Arg(z) + 2k\pi)$$

An easy example is $z = 1$, which gives $log(z) = 0$ and $log(-z) = i\pi$. So $log(z) \ne log(-z)$ in the general case.

Am I missing something obvious here? Thanks for any pointer.

Answer 1

Note that $|1-2e^{it}| = |-1+2e^{it}|$. So if there's any difference in the two expressions, it's in the argument. The argument of $1-2e^{it}$ is equal to $\pi$ plus the argument of $-1+2e^{it}$, since it is the opposite vector. This $\pi$ (times $i$, from $i(\arg z+2k\pi)$) is absorbed into the "+constant."

Answer 2

Recall that $\log(r e^{i\theta})=\log(r)+i(\theta +2n\pi)$, where $n$ is an integer. Also $e^{i\pi}=-1$, so $$\log(-z)=\log(z e^{i\pi})=\log(r e^{i\theta+\pi})=\log(r)+i((2n+1)\pi+\theta)=\log(z)+i\pi$$

So it appears the confusion arising from wolfram alpha's expression is the $+$ constant term.