I was thinking about branch cuts in the following way.
Suppose $f(z)= (z^2-1)^{1/2}$. Then, I was taught that this is the same as $$ e^{1/2 (\log (z+1) + \log (z-1)}. $$ Then it suffices to pick branches for $\log z$ and $\log (z-1)$ and then combine them to get a branch for $f$. My issue with that is that I am not convinced that $$ f(z)=e^{1/2 (\log (z+1) + \log (z-1)}. $$ This is because $(ab)^{1/2} \neq a^{1/2} b^{1/2}$ for all $a$ and $b$. However, this logic seems to be used through, or I am simply not understanding the argument.
There have been a lot of questions on this website about branch points, however, I do not think they address my concern here.
Questions:
- Is there any error in the above logic?
- How can I define $f$ via the logarithm function?
- With a given branch selection how does one actually evaluate $f$ at a given point?
Remarks: I am aware that a simple way to get around the logarithms is just to use integrals, however, I would like to understand how one does this with $\log$.
I decided to rewrite another answer because I think I was not dealing with your question properly.
Firt of all, it is true that the statment
$$ (ab)^{\frac{1}{2}} = a^{\frac{1}{2}}b^{\frac{1}{2}}$$
is not valid in general. This happen when we restrict the logarithm function with a branch:
Let $z=re^{i\Theta}$. The general definition of the $\ln(z)$ function over the complex plane is:
$$ \ln(z) = \ln(r) + i\left(\Theta+2\pi n\right) \quad n=0,\pm 1, \pm 2, ..$$
Now if $z_{1}= r_{1}e^{i\theta_{1}}$ and $z_{2}= r_{2}e^{i\theta_{2}} $
we have
\begin{align*} \ln(z_{1}z_{2}) = &\ln\left(r_{1}r_{2}e^{i(\theta_{1}+\theta_{2} +2\pi k )}\right) \quad k=0,\pm 1, \pm 2,.. \\ = &\ln(r_{1})+\ln(r_{2}) + i(\theta_{1}+\theta_{2} +2\pi k )\\ \end{align*}
On the other hand,
$$ \ln(z_{1}) = \ln(r_{1})+i(\theta_{1} +2\pi n) \quad n=0,\pm 1, \pm 2,..$$ $$ \ln(z_{2}) = \ln(r_{2})+i(\theta_{2} +2\pi j) \quad j=0,\pm 1, \pm 2,..$$
So we could always find $k,n,j$ where
$$ \ln(z_{1}z_{2}) = \ln(z_{1})+\ln(z_{2}) \tag{1} $$
holds. This is no longer true when we define a branch for the logarithm. Let $\alpha \in \mathbb{R}$, then:
$$ \ln(z) = \ln(r) + i\Theta \quad \alpha < \Theta \leq \alpha + 2\pi $$ this definition makes the function single-valued. If we restrict the range to $\alpha < \Theta < \alpha + 2\pi$ the function will also be analytic. If we put $\alpha = -\pi$ this interval is called the principal branch
Is easy to see that the restricted $\ln$ does not satisfy the product property; for example, take $z=-1$, $\ln((-1)(-1)) =0$ while $\ln(-1) +\ln(-1) = 2i\pi$.
Now, for your problem:
From property (1), if we leave the $\ln$ function unrestricted we have:
$$ (z^2-1)^{\frac{1}{2}} = e^{\frac{1}{2}\ln(z+1)+ \frac{1}{2}\ln(z-1)} = (z+1)^{\frac{1}{2}}(z-1)^{\frac{1}{2}}$$
This is perfectly valid. However, we can mantain the equality even if we restrict the branches of $(1+z)^{\frac{1}{2}}$ and $(z-1)^{\frac{1}{2}}$. The trick is to choose a branch for the left hand side, $(z^2-1)^{\frac{1}{2}}$, carefully:
Let $$r_{1}=|z-1| \textrm{ and } \theta_{1}=\arg(z-1)$$ $$r_{2}=|z+1| \textrm{ and } \theta_{1}=\arg(z+1)$$ restrict the arguments to these branches:
$$ (z-1)^{\frac{1}{2}} = r_{1}e^{\frac{i\theta_{1}}{2}} \quad r_{1}>0, 0<\theta_{1}<2\pi \tag{2}$$
$$ (z+1)^{\frac{1}{2}} = r_{2}e^{\frac{i\theta_{2}}{2}} \quad r_{2}>0, 0<\theta_{1}<2\pi \tag{3} $$ Then
$$(z+1)^{\frac{1}{2}}(z-1)^{\frac{1}{2}} = \sqrt{r_{1}r_{2}}e^{\frac{i(\theta_{1}+\theta_{2})}{2}} \quad r_{1}>0, r_{2}>0, 0<\theta_{k}<2\pi, k=1,2 \tag{4} $$
On the other hand:
$$(z^2-1)^{\frac{1}{2}} = e^{\frac{1}{2}\ln\left(r_{1}r_{2}e^{i(\theta_{1}+\theta_{2})}\right)} = e^{\frac{1}{2}\ln(r_{1}r_{2}) + \frac{1}{2}i(\theta_{1}+\theta_{2}+2\pi n)} =\sqrt{r_{1}r_{2}}e^{\frac{i(\theta_{2}+\theta_{2}+2\pi n)}{2}} \tag{5} $$
So, the product of the two branches in (2) and (3) define a branch for (5) given by the conditions in (4) (ie $n=0$). As long as we stick to these conditions the functions should be equal.
Note that the function in $(4)$ is defined everywhere in the $z$ plane except on the half line $\Re(z)\geq -1, \Im(z)=0$.