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I'm currently following a class on complex analysis (for physicists) and I'm having trouble with the concept of branch points. In my textbook, they say that log(z) has a branch point at zero and at infinity. They also state that every other point is not a branch point and i'm having a hard time proving that $z=2$ is actually not a branch point. I've searched a lot without finding anything convincing.

So far I tried the following :

Let's say we look at the function $f(z) = ln(z)$ at $z=2$. To show that this is, or is not a branch point, we can draw a circle of radius 1 around this point. Thus, let us take $z_i = 2+e^{i\cdot 0}$ and $z_f=2+e^{2\pi i}$. In that case, $f(z_i)=\ln(2+1) = \ln(3e^{i2\pi n})$ and $f(z_f)=\ln(2+1) = \ln(3e^{i 2\pi n})$ and thus $ f(z_i)=f(z_f)$, which is what we want. However, if I take a circle of radius 3, the result of this methode is the sames, altough it shouldn't since the circle goes around zero. Indeed, in that case, this method leads to :

$f(z_i)=\ln(2+3) = \ln(5e^{i2\pi n})$ and $f(z_f)=\ln(2+3) = \ln(5e^{i2\pi n})$...

I guess there is something wrong with what I'm doing, but I don't know what...

I would like to show that $z=2$ is not a branch point using contours, not by other means.

Thanks a lot!

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    If $f(z) = e^z$, then $f(\log 2) = 2$, and $f'(\log 2) \neq 0$, hence $f$ has a local, analytic inverse $g$. It follows that $\log$ does not have branch point at $2$. – 2017-02-18
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    Is there a way to find this answer using contours? – 2017-02-18
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    I don't understand what you mean by 'using contours'. If $g$ (which is $log$ plus a constant) is analytic at $2$ then the value of $g$ locally is independent of any path. – 2017-02-18
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    Page 4 of that textbook : (http://math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf) states that ''consider again the case of the function log(z), take z0 = 2 and a closed curve around z = 2 that also encloses z = 0. The value of log(z) wil l change as this curve is traced, but this does not make z = 2 a branch point. In fact, for curves close to z = 2 there is no change, hence z = 2 is not a branch point of log(z)!) '' – 2017-02-18
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    The inverse function theorem shows that $\log$ is analytic in a convex neighbourhood of $2$. So there is no issue with definition. If you take a path around $2$ that also encloses $0$, then the function will need to be extended ('define it continuously' in the book) as one goes around the path and on return to the starting point will have accumulated some multiple of $2 \pi$ (depending on the net encirclements). The function $\log$ is defined everywhere in a neighbourhood of $2$ so there is no need to extend. – 2017-02-18

2 Answers 2

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Bit late to this one, but if it's wrong someone let me know. When we say a branch point it means if we take an arbitrary path in the z-plane, following this path in the w-plane from the continuous mapping of w = f(z). Then the same value for f(z) should be retained.

In the example: $$f(z) = ln(z),$$ at the point z = 2 we can consider the path as a circle of radius R. Therefore :

$$z = 2 + Re^{i\theta}$$

So $$ln(z) = ln(2 + Re^{i\theta})$$

so for z = 2 to be a branch point, cycling through values of $\theta = 2\pi k$, where k is an integer, we must get different values for $ln(z)$, which we don't.

Considering a known branch point, $z = 0$, we get the expression: $$ln(z) = ln(0 + Re^{i\theta})$$ $$ -> ln(z) = ln(R) + i\theta$$

Which clearly changes as $\theta$ increases from $0$ to $2 \pi$. So $z = 0$ is a branch point.

Hope this is correct.

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You need that the function can be defined like singlevalue function and this definition must be at least continuos inside the set joined to their interior. This assumption doesn't satisfy in the second contour.