branch-cutsbranch-pointsbranching-rulescomplex-analysis
I am reading a complex analysis book. I am trying to understand if there is an algorithmic/procedural way to identify branch points and branch cuts.
For example, I have the function $ln(1-z^{1/2})$.
I know that $ln(0)$ is undefined. I set $1-z^{1/2}=0 \rightarrow z=1$. Does this mean that 1 is a branch point? Is it the only branch point?
Does this mean that the branch cut is $(-\infty,1) \cup (1, \infty)$?
If not, how do I find the branch cut?
Thank you very much.
With regard to 1) Calculus may shed a little light. If $f(z)=\sqrt{z}$, then $f'(z)=\dfrac{1}{2\sqrt{z}}$, and we see that $f'(z)$ is undefined at $z=0$ (where $f$ vanished). In terms of real variables the derivative doesn't exist at $0$, but in terms of complex variables $0$ is a non-isolated singularity, i.e. $f(z)$ is not analytic in a deleted neighborhood of $0$. The same holds true for $f(z)=\sqrt[n]{z}$.
It seems like 2) may indicate some misunderstanding related to an answer to 3). Geometrically a branch cut is a cut, a disconnection in the plane usually between pairs of branch points. It's as if a razor had been used. A richer geometric picture emerges from considering an alternative approach to branch cuts for handling branch points of multi-valued functions called Riemann Surfaces.
The encircling of branch points in the linked answer you referenced is a standard approach to determining branch cuts. The issue we must avoid is a discontinuity of the function after making a complete circuit around a point. We need to ensure that traversing a closed loop produces a change in the argument of $f(z)$ which is a multiple of $2\pi$. The simplest example is $f(z)=\sqrt{z}$. A closed loop around $z=0$ produces a change in the argument of $f(z)$ by $\pi$ for a change of $2\pi$ in the argument of $z$. That is what indicates that a branch cut from $0$ to $\infty$ is required. A closed loop around $0$ cannot be allowed, because it's impossible to avoid a discontinuity of $f(z)$. (Think about the loop $e^{it}$. At $t=0, z=1, \sqrt{1}=1\text{ and }\arg f(z) = 0$, but after a circuit around $z=0$ at $t=2\pi, z=1\text{ again, but, } \sqrt{1}=-1\text{ since}\arg f(z) = \pi$.
What the linked answer shows is that a loop around either $z=i$ or $-i$ alone produces a change of $\pi$ in $\arg \sqrt{z^2+1}$ which would be a problem, but a loop enclosing both of the points produces a change of $2 \pi$ in $\arg \sqrt{z^2+1}$. A branch cut joining $z=i$ and $z=-i$ ensures that any loop enclosing one of the points must enclose the other.
(1)
To say that the branches of the factors cancel is to say that $f$ is still analytic along the overlap of the branches (excluding the endpoint(s) of the overlap).
Branches do not always cancel where they overlap: For example, we can generalize $f$ to
$$f_k(z) := \sqrt[k]{z - 1} \cdot \sqrt[k]{z + 1}$$
for positive integers $k$. The locations of the branch cuts of the two radical expressions (making the same choice of logarithm branch as in the question) are the same as for $f$, but the branch cuts do not "cancel" on their overlap for $k \geq 3$. For $\rho > 1$ and small $\epsilon$, $$f_k(-\rho e^{(\pi \mp \epsilon) i}) = e^{\pm 2 \pi i / k} \sqrt[k]{\rho^2 - 1} + O(\epsilon) .$$ In particular, $f_k$ is only continuous at $-\rho$ for $k = 1, 2$.
Plot of $f_3$; note the discontinuity in the argument along $\{z < -1\}$.
$f_3$">
Roughly speaking, checking continuity at an arbitrary point of the cut (excluding the branch point) is generally sufficient to establish (non)cancellation: Analyticity of the function then follows from the Inverse Function Theorem.
(2) I'm not sure how to make your question precise.
(3) This is certainly not true in general. For example, if $h$ is the zero function, so is $g = h \circ r$, but this function has no branch points. In general, the only way that $h \circ r$ can avoid inheriting a branch cut from $r$ is, roughly speaking, for $h$ to map the distant points to which $r$ maps nearby points on opposite sides of the branch cuts back to nearby points. So, continuing our example from before, taking $r = f_k$ and $h(z) := z^k$, we have $$g(-\rho e^{(\pi \mp \epsilon)}) = h\left(e^{\pm 2 \pi i / k} \sqrt[k]{\rho^2 - 1} + O(\epsilon)\right) = \left(e^{\pm 2 \pi i / k} \sqrt[k]{\rho^2 - 1} + O(\epsilon)\right)^k = \rho^2 - 1 + O(\epsilon) ,$$ so $g$ is continuous on $\{z < -1\}$, and a similar argument shows that $g$ is continuous on $\{-1 \leq z \leq 1\}$, too; in fact, $g(z) = z^2 - 1$.
(4) No. For $z = -i$, for example, $\sqrt{z - 1} \sqrt{z + 1} = -\sqrt{2} i$, but $\sqrt{z^2 - 1} = \sqrt{2} i$.
Best Answer
We consider two functions $w=1-z^{1/2}$ and $\log w$.
We define normally $$z^{1/2}=\sqrt{r}e^{i\theta /2} \quad(z=re^{i\theta },\, |\theta |<\pi), $$ so $z^{1/2}$ is defined for $z\in \mathbb{C}\setminus (-\infty,0\,]$ and $(-\infty,0\,]$ is the branch cut of $z^{1/2}$.
Then $z^{1/2}$ satisfies $\operatorname{Re }z^{1/2}>0$ since $|\arg (z^{1/2})|<\frac{\pi}{2}.$ From this we see $$ \operatorname{Re} w=\operatorname{Re }\left(1-z^{1/2}\right)<1.$$ On the other hand $\log w$ is defined normally as follows (the principal branch):$$ \log w=\log r+i\theta\quad (w=re^{i\theta }, |\theta |<\pi).$$ In other words, $\log w$ has a branch cut $(-\infty,0\,]$. We note that $-\infty<w=1-z^{1/2}\le 0$ corresponds to $1\le z^{1/2}$, which corresponds to $1\le z.$
Thus $\log (1-z^{1/2})$ has a branch cut $(-\infty, 0\,]\cup [\,1,\infty)$.