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I'm a little confused about the branch cut thing. Given an entire functions $f(z),g(z),h(z)$, $z\in \mathbb C$, such that $f(x)=g(x)+h(x)$ for all $x\in \mathbb R$, $f$ and $g+h$ doesn't vanish on $\mathbb R$ . I take the $\log$ for both sides then differentiate and get:

$$\log f(x)=\log(g(x)+h(x))$$ $$\frac{f'(x)}{f(x)}=\frac{g'(x)+h'(x)}{g(x)+h(x)}$$

I thought this is correct, but I was asked about "what is the branch cut I used here?", and I don't know what does this mean, and what is the answer for this question! Any help!?

Edit: I also have the same problem with the following case:

If $f(x)=e^{g(x)}$ then $\log f(x)=g(x)$, and $\frac{f'(x)}{f(x)}=g'(x)$, also a branch cut issue!

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If $f=k$ and $f$ doesn't vanish then it is true that $f'/f = k'/k$, simply because $f=k$ and $f'=k'$; no logs needed. It is also true that whenever a logarithm of $f$ exists, i.e. a function $p$ such that $\exp(p)=f$, then it is true that $p'=f'/f$, a consequence of the chain rule. I'm not sure exactly what the question is here. Maybe you should ask the person who asked you for clarification of what they are asking. They might just be hinting at the fact that logarithms don't always exist globally and aren't unique, so care should be taken in what "log" means. – Jonas Meyer Dec 4 '11 at 20:33
One thing to note: There is no continuous logarithm defined on the range of a nonconstant entire function. However, for every nonvanishing analytic $f:U\to \mathbb C$, where $U$ is simply connected, there exists an analytic $p:U\to \mathbb C$ such that $\exp(p)=f$; such a $p$ is called a logarithm of $f$. – Jonas Meyer Dec 4 '11 at 20:46

Sometimes, problems arise when trying to extend continuous functions from $\mathbb{R}$ to $\mathbb{C}$. For example, consider the complex-valued function $f(z) = \sqrt{z}$. Consider the following (poorly-drawn) diagram:

Complex Square Root

If your domain $D$ sweeps out a quarter-circle of radius $r$, the the image $f(D)$ sweeps out half the angle (an eighth-circle) with radius $\sqrt{r}$. Suppose then that your domain $D$ was an entire circle in the domain.

enter image description here

Consider the blue-red interface in the domain. As the image suggests, $\sqrt{z}$ is discontinuous there, because as you transition across the line $x = 0$ in the domain, you jump half-way across the circle in the image. To deal with this, we extend $\sqrt{z}$ to a new domain: two complex planes "glued" together. It looks something like this.

enter image description here

Now the function is continuous. Instead of transitioning directly from (1) to (2) in the image plane, we "move through the cut" to the blue (3) on the second complex plane, so that $\sqrt{z}$ is continuous. Then we travel around the second plane, jumping from the red (4) to the red (2) in the first complex. We have gone around twice in the domain, and once in the image, and our issues with continuity are resolved. To be more precise about the gluing, the domain actually looks like this:

enter image description here

Now, as you can expect, you need to glue three complex planes together to make $\sqrt[3]{z}$ continuous, and so on, etc. Unfortunately, for the complex logarithm, no number of complex planes in the domain will make the image continuous. Thus, you need to define the complex logarithm on an infinite collection of complex planes, "spiraling" away in either direction. It looks like this.

enter image description here

Recall the example with the square root earlier. We had two copies of the complex plane, and their images under $\sqrt{z}$ were different. In the first copy, the image was the right semi-circle, and in the second copy, it was the left semicircle. Thus, when working with the complex square root, one needs to specify which copy of the complex plane one wishes to work with. Each such copy is called a branch (symmetrically, you can think of it as two copies of $\sqrt{z}$ defined on the same complex plane). Similarly, when working with the complex log, you need to talk about which of the infinitely many complex planes in the domain you wish to work with, and so you must specify which branch you are using. The popular choice is the so-called "principal branch":

Log$(z) = ln|z| + i$Arg$(z)$.

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...very nice pictures, btw – Grigory M Dec 4 '11 at 21:19
Those are nice Riemann surfaces you've shown... :) – J. M. Dec 5 '11 at 1:10
Wow! Nice pictures indeed. – Nick Strehlke Feb 1 '12 at 7:42

The problem here is that $\log(z)$ is a multi-valued function on the complex plane. How can you see this? First of all recall that $\log(z)$ is usually defined to be an inverse of $e^{z}$ (i.e. $log(e^{z})=z$). Working out what this means shows that one way of defining the $\log$ is to take $$\log(z)=\ln|z| +iArg(z)$$ but then you run into the problem that $Arg(z)$ has an infinite number of values, just add or subtract a multiple of $2 \pi$. Now consider $\log(1)$. We want $\log$ to be continuous (and even analytic), so it should satisfy $$\lim_{y \to 0^{+}} \log(1+iy) =\lim_{\theta \to 2\pi^{-}} \log(1e^{i\theta}) = 0$$ The first one appears to be true, but the above definition of log shows that in the second limit $Arg(z)$ approaches $2\pi$ so that $$\lim_{\theta \to 2\pi^{-}} \log(1e^{i\theta}) = 2\pi i$$ The problem here is that the domain of the log function is not simply connected. A way to fix this problem is to remove all of the negative real axis and zero so that you can't go all the way around $0$ and force 2 different values of $Arg(z)$. Removing these points gives you an analytic function on $\mathbb{C}\setminus \{x\in \mathbb{R}|x\leq 0\}$ and this is called a branch cut because it singles out a single valued function of $\log$ from an infinite number of choices for the range of $Arg(z)$. One can also make a branch cut by removing any ray that connects zero and infinity.

In your problem, we are not sure of what domain your final statement is true on as we dont know what branch of the $\log$ function you have chosen.

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OH guys, your making things more difficult! – Katie Ml Dec 4 '11 at 22:11

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