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I'm starting to learn a little complex analysis, and I'm a little confused as to what the purpose of a branch cut is. Is it to make a function continuous, or single valued?

For example, the $\sqrt{}$ function is multivalued: since $w(z)=z^2$ has $w(z)=w(-z)$, in order to make $\sqrt{}$ the inverse of $z^2$, we can choose one of two branches, $w\to\sqrt{w}$ or $w\to -\sqrt{w}$, the former being the principal branch. This function maps $\mathbb{C}\to$(right half plane), and the other maps to the left half plane. As far as I can tell, both functions are now well defined on $\mathbb{C}$, but are not continuous because for a point $-r$ on the negative real axis, the principal branch $w(z)\to i\sqrt{r}$ as $z\to r$ from above, but $w(z)\to -i\sqrt{r}$ as $z\to r$ from below. To make the principal branch continuous, we just cut out this section of the real line and define the principal branch $\sqrt{}_{+}:\mathbb{C}\setminus (-\infty,0]\to\mathbb{C}$.

Is this the right idea? Also, I was assuming Arg$(z)\in[-\pi,\pi)$; is this the only branch cut you could make with Arg$(z)\in[-\pi,\pi)$? Where would the branch cut be if Arg$(z)\in[0,2\pi)$ instead?

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  • $\begingroup$ There are an infinite number of possible branch cuts that one can choose such that $z^{1/2}$ is analytic in the plane less the cut. If one cuts along the positive real axis, then $0\le \arg(z)<2\pi$. And if one cuts along the line $y=x, x>0$, then one has $- 3\pi/4 <z\le \pi/4$ for example. And lastly, the branch cut need not take a linear path. $\endgroup$ – Mark Viola Aug 23 '15 at 21:07
  • $\begingroup$ I had read elsewhere that any curve that isn't self-intersecting will work for a branch cut. You said that if one cuts across the positive real axis, then $0\leq arg(z)<2\pi$, but is this the only cut you can make for this particular restriction for the values of $arg(z)$? $\endgroup$ – user153582 Aug 23 '15 at 21:16
  • $\begingroup$ If we measure the argument of $z$ as the angle relative to the positive $x$ axis, then can their be another cut for which $0\le \arg (z) < 2\pi$? $\endgroup$ – Mark Viola Aug 23 '15 at 21:21
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It is to make the function continuous AND single valued. In modern mathematics, "multiple-valued" are not accepted as functions. One important reason is that it is not clear how to add or multiply them.

Consider your example $z^2=w$. This is an equation, $z$ is unknown, and $w$ is a parameter. When $w\neq 0$ this equation has two solutions. So if we want to define a function $z=f(w)$ solving this equation, and giving all solutions, it has to be $2$-valued, and this is not permitted, as I said.

We can try to SELECT one of these two solutions, for each $w$, and make a function $z=f(w)$ which will satisfy the equation that is $f^2(w)=w$ for all $w$. The question is how to select.

This can be done in VERY many ways (infinitely many). In real analysis, they usually use positive $w$ then there is a natural way to select: you take as $f(w)$ the positive solution of the equation $z^2=w$. With complex numbers this does not work.

One may want to select $f(w)$ is a CONTINUOUS way. This is not possible if we want to define $f$ for all $w$. Prove this yourself. This is why we need to make a branch cut. If we remove the cut (for example along the negative real axis) from the plane, then a continuous selection becomes possible in the remaining region. Still it is possible in two ways. Choosing $f(1)=1$ makes a continuous selection unique. This $f$ is called the principle branch. In the same region there is another branch, namely $-f$. But if you try to define them on the cut, you see that this is impossible if you want to keep them continuous. Simply because the limit from above of the cut is not equal to the limit from below of the cut. And this is for each of the two "branches", $f$ and $-f$.

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