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I am trying to find a function where multiple branch cuts are required.

One example I've encountered online was $\sqrt{(z)}\sqrt{(z-1)}$; however, I am confused as to why the branch cut $y = 0, x \leq 1$ and $y = 0, x \geq 1$ prevents encirclement of just one branch point. Does this prevent encirclement of the branch point at $z = 0$ or just $z = 1$ and how so, how does that give us a branch of analyticity.

I am also trying to figure out how verify the branch point at z = 1 as the angles change from $\theta_1 = 0$ and $\theta = 0$, to $\theta_1 = 2\pi$ and $\theta = 0$ Textbook

I also am wondering if you could provide me with a function that has multiple branch cuts, that we can illustrate without encirclement of the function, and can you please help explaining how those branch cuts help create at least one analytic branch.

So there a three question in total, but the main idea (in bold) of showing how multiple branch cuts can create analyticity is what I am after

Thanks

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The point of the branch cut is to limit the range of values for the phase of the complex number below the square root. Around $0$, write $z=r e^{i \theta}$, $\theta$ is defined up to a multiple of $2\pi$ so in fact we should write: $z=r e^{i (\theta+2k\pi)}$ with integer $k$ and say $\theta \in [-\pi,\pi[$. Now take the square root bluntly: $\sqrt{z}=\sqrt{r} e^{i (\theta/2+k\pi)}$: the term $e^{ik\pi}$ contributes a $\pm$ sign according to the parity of $k$ which is only natural since even for real numbers the square root is defined up to a sign. In order to define a continuous function $z \in \mathbb{C} \to \sqrt{z}$ you need to limit the allowed values of $k$. This is the role of the branch cut. If you remove say the negative real axis from the complex plane, then you can define a single valued phase for each $z \in \mathbb{C} \setminus \mathbb{R}^-$ by taking $\theta \in ]-\pi,\pi[$. Now the square root is single valued in $\mathbb{C} \setminus \mathbb{R}^-$. The choice of the negative real axis is not unique and any half-line starting from $0$ will do. In fact any simple curve $\Gamma$ starting from $0$ such that $\mathbb{C}\setminus \Gamma$ will work also. If you consider $\sqrt{z-1}$, the branch cut should start from $z=1$. Considering your function $\sqrt{z}\sqrt{z-1}$, you have to make sure that the phase of $z$ and $z-1$ are uniquely defined. This means that you are not allowed to add $2\pi$ to the phase of $z$ and $z-1$. This can be done by taking two parallel half-lines starting from $z=0$ and $z=1$, or to draw a line between $z=0$ and $z=1$ or any simple curve joining $z=0$ and $z=1$.

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  • $\begingroup$ Can you further explain why the give choice $y = 0, x \leq 0$, and $y = 0, x\geq 1$ make it analytic at least at one branch, do you know if that is relating to the branch z = 0 or z = 1. Also, do you have another example for multiple branch cut function, I was reading that $e^{(1/2)}$$\log\sqrt{(z^2-1)}$ has branch cuts at $-1 < x < 1$ for the real part and $ \inf \leq y \leq -\inf $ for the imaginary part. Why is that the case? $\endgroup$
    – Elli
    May 6, 2021 at 6:47
  • $\begingroup$ $y=0, x \leq 0$ is a half-line starting from$z=0$ and the other one starts from $z=1$. You should make a drawing. I do not understand the definition of the new function (it looks like a constant times a $\log$) neither your definition of the branch cut, from what you wrote the "branch cut" is an infinite band. $\endgroup$ May 6, 2021 at 7:34
  • $\begingroup$ The new function has constant $(e^(1/2))$ in front of log and I'm wondering firstly if this has multiple branch cuts (it seems like it does), and two if you are able to explain to me the nature of these branch cuts and why they are made, removing the discontinuities $\endgroup$
    – Elli
    May 6, 2021 at 19:56
  • $\begingroup$ It is a bit long to answer that, this is not the purpose of comments. Accept the answer and ask another question. $\endgroup$ May 6, 2021 at 21:38

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