The function $\log(z)=\log|z|+i \arg(z)$, is continuous in the domain $\mathbb{C}-\mathbb{R^-}$.
But can we say that it's defined on $\mathbb{C^*}$.
Like for example : $\log(-1)=\log(\exp(i\pi))=i\pi.$ is it true ?
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$\begingroup$ Or maybe $\log\left(-1\right)=3\pi$ since $e^{3\pi i}=-1$. What would you choose? $\endgroup$– eranrechesCommented Jun 6, 2019 at 22:15
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$\begingroup$ Do you mean that $\log(-1)$ can equal $3\pi i$? $\endgroup$– mjwCommented Jun 6, 2019 at 22:20
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$\begingroup$ Yes. In complex analysis some function are multi-valued and it is up to you to choose one out of all. $\endgroup$– eranrechesCommented Jun 6, 2019 at 22:29
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2 Answers
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Well, you can either choose your branch cut somewhere else (for example, along the negative imaginary axis) or you can form a Riemann surface (then the function $f(z)=\log z$ becomes multiple valued). In theses cases, the function is defined along the negative real axis.
If your branch cut is along the negative real axis, then your function is not defined there.
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$\begingroup$ So for example taking the branch $(-\pi,\pi]$ is good, but $(0,2\pi]$ is not ? $\endgroup$– BrianTagCommented Jun 6, 2019 at 22:39
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$\begingroup$ Either branch, or any line or curve joining the origin to the point at infinity is a valid branch cut. Points along the branch are undefined. If you require the function to be defined at a particular point, then you would need to define a branch cut that does not go through it. $\endgroup$– mjwCommented Jun 7, 2019 at 2:26
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Before I answer your question, I'll make the note that you are writing $\log(z)$ on the left side of your equations, but are returning the principal value of the $\log(z)$ function on the right side. The principal value has the branch cut, $\log(z)$ does not. This was denoted in my textbook as $\text{Log}(z)$, or $p.v. \log(z)$.
Even the principal value is defined everywhere on $(-\pi, \pi]$. So yes, you are correct $\text{Log}(-1)=i\pi$. The problem arises when you try to integrate the value. Because,
$$\lim_{\epsilon\to0}\left(\text{Log}(-1-i\epsilon)\right)=-i\pi$$
which is off by $2i\pi$ from the result of $\text{Log}(-1)$. This is problematic when we try to integrate, so we cannot integrate through this branch cut. However, the function is defined on the negative real axis, it just becomes discontinuous.
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1$\begingroup$ Yes, thank you for the clarification, you answered my question, I knew that $Log$ is discontinuous on the negative real axis, but it's defined! $\endgroup$– BrianTagCommented Jun 6, 2019 at 22:37