Talk:Branch point

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A point about the now-changed formula

F(z) = \sqrt{z} \sqrt{1-z}

is that it is no innocent difference, to take the square root of each factor separately; each then has a different domain of definition, so that F is actually defined a priori on the intersection of the domains. Which is to confuse the point being made with the example, really.

Charles Matthews 11:05, 15 Oct 2004 (UTC)

[edit] Mistake

Is the statement:

A point z0 is a branch point for a holomorphic function f(z) if and only if its derivative f ′(z) has z0 as a simple pole (i.e., a pole of order 1) − see mathematical singularity.

correct? The function \sqrt{z} seems to be a counter example. 0 is a branch point, but its derivative is 1/2\sqrt{z} which also has a branch point at zero, not a pole of order 1.

It's a strange thing to say. Branch points basically are places where the inverse function theorem fails, in this setting. So they are associated with zeroes of the derivative of the function to be inverted. Charles Matthews 09:55, 16 Jan 2005 (UTC)
That's funny, because I found your statement strange. :) I see your point, and I understand what you are trying to say. However, let us be rigurous.
  • A branch point is not a place where the inverse function theorem fails. For the function z->z2 the inverse function theorem fails at 0, but 0 is in no way a branch point for this function. A branch point, according to what the article says, is a point where the function is multi-valued, like log z at 0.
  • The inverse of z->z2, which is the square root, does have 0 as a branch point. However, the derivative of the square root does not have 0 as a pole.

In short, there is some connection between branch points and poles, but the statement

A point z0 is a branch point for a holomorphic function f(z) if and only if its derivative f ′(z) has z0 as a simple pole (i.e., a pole of order 1) − see mathematical singularity.

is still wrong. I am talking wrong mathematically, that statement cannot be proved to be right, because I gave a counterexample (the second item above). Oleg Alexandrov 17:23, 16 Jan 2005 (UTC)

You are correct, of course. I was trying to put the ramification concept into quite elementary language. Charles Matthews 19:18, 16 Jan 2005 (UTC)

[edit] ?

perhaps a definition? —The preceding unsigned comment was added by Rsjyufoih (talk • contribs) .

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