2
$\begingroup$

Consider the function $$f(z) = \ln (z^2-1),$$ which as we know is a multi-valued function. Now doing $$f(z)=\ln[(z-1)(z+1)]=\ln(r_1r_2 e^{i(\theta_1+\theta_2)}),$$ where I let $z-1=r_1e^{\theta_1}$, $z+1=r_2e^{\theta_2}$.

To make single-valued, I let $$-\frac{3\pi}{2}<\theta_1\leq \frac{\pi}{2}, \, -\frac{3\pi}{2}<\theta_2\leq \frac{\pi}{2}.$$ To find its branch cut, I need to find where $f(z)$ is discontinuous. So $$-3\pi<\theta_1+\theta_2\leq \pi \, \Rightarrow \, -\pi<\theta_1+\theta_2\leq \pi.$$ So the discontinuous location should be the negative real axis.

Am I making any sense? Thanks!

$\endgroup$
1
1
$\begingroup$

In case this link Defining a holomorphic branch of $\log(z^2-1)$ does not address your specific question:

If you simply cut along the negative real axis, the resulting domain will still include the point $z=+1$ at which no branch of $\log(z^2-1)$ can be defined. Any cut must eliminate bot $z=1$ and $z=-1$.

But there is another way to approach obtaining a single valued holomorphic version of $\log f(z)$, where for simplicity we will take $f(z)$ to be holomorphic except at a finite number of zeros and / or poles. We can require a domain $\Omega$ in which every path $\gamma \subset \Omega$ from $z_0$ to $z$ gives the same value, $$ \log z = \log(z_0) + \int_\gamma \frac{f'(z)}{f(z)} ~ dz. $$ This can only arise if all paths joining $z_0$ to $z$ are homotopic in $\Omega$, that is each can de deformed continuously into the other without encountering poles of $f'(z)/f(z)$.

That will be the case if, as above, the domain excludes all the zeros of $f(z)$ and the paths $\gamma$ cannot circulate the poles of $f'(z)/f(z)$. There may be exceptional cases where the sum of their residues is zero, and then there may be more flexibility, but we will momentarily exclude this case. Then, to avoid circulation, a sufficient condition is that the domain be simply connected.

In our case $f(z) = z^2-1$ and the residue sum for $f'(z)/f(z)$ is non-zero, so our cut must prevent a circuit to include $z =\pm 1$. That still leaves many choices. One is the domain that is cut along the real axis from $z=-\infty$ to $z=1$, so that $\Omega = \mathbb C \setminus (-\infty, 1]$. Another, used in the link, is to take $\Omega = \mathbb C \setminus \Big( (-\infty, -] \cup [1,\infty) \Big).$ Both give different holomorphic versions of the $\log$ function.

Any cut can be made to work if it meets these criteria: choose the domain $\Omega$, take a point $z_0 \in \Omega$ and its corresponding value $\log z_0$ and define, $$ \log(z) = \log(z_0) + \int_{z_0}^z \frac{2w}{w^2-1}~dw = \log(z_0) + \int_{z_0}^z \frac{1}{w+1} + \frac{1}{w-1}~dw. $$ This will give a single valued holomorphic branch. The value is independent of the path by Cauchy's theorem.

Coming back to an example where the residues do cancel, consider $\displaystyle \log\frac{z-1}{z+1}$.

Here $f'(z)/f(z) = 2/(z^2-1)$, the residues are $\pm1$ and the corresponding cut for $\log z$ need only exclude jointly the points $z = \pm 1$. A commonly used domain in this case is $\Omega = \mathbb C \setminus [-1,1]$, although you could again use $\Omega = \mathbb C \setminus \Big( (-\infty,-1] \cup [1,\infty) \Big)$.


Addition: I should have mentioned that for $f'(z)/f(z)$ every zero of $f$ has residue 1 and every pole residue -1, so the sum of the residues is simply the number of zeros less the number of poles.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.