I have a function $f(z) = (\frac{8}{7}z^3-\frac{64}{7})^\frac{1}{3}$ which I have found to have branch points at $z_1 = 2$, $z_2 = 2\exp(\frac{2i\pi}{3})$, and $z_3 = 2\exp(\frac{-2i\pi}{3})$. The part I am stuck on I am suppose to consider branch cuts from $z_1$ to $z_2$. Then, let $F(z)$ denote the branch when Im[$f(10)$] = 0. What is the value of $F(1)$. I'm confused on how exactly I find $F(z)$.
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Before you tackle this problem, do you understand the simpler case $g(z) = z^{1/3}$? The branch points are 0 and $\infty$. If Im[$g(1)$] = 0, what is $g(-1)$ if you consider a branch cut from 0 to $\infty$ going upwards along the imaginary axis? What is $g(-1)$ if you consider a branch cut from 0 to $\infty$ going downwards along the imaginary axis? – Ted Feb 14 '14 at 05:37
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I don't really know how to even this simpler case. For the case with a branch cut from 0 to $\infty$ upwards along the imaginary axis would I first evaluate $g(z)$ at 1, resulting in $g(1) = 1$. Then the Im[$g(1)$] = 0? I really am not sure how to approach it. – Jim Feb 14 '14 at 05:56
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Before we tackle this problem, consider the simpler case $g(z)=z^{1/3}$, with branch points at 0 and $\infty$. Suppose Im[$g(1)$] = 0. For each of the possible branch cuts joining 0 and $\infty$, what is $g(-1)$? Clearly $g(-1)$ has to be one of the 3 cube roots of -1. The three cube roots of -1 are -1, $e^{-\pi i/3}$, and $e^{\pi i/3}$. So which one is it?
The main reason that branch cuts are necessary for $g(z) = z^{1/3}$ is because the arg function $\theta$ cannot be defined continuously for all nonzero $z$. If we could define $\theta$ continuously, then we could take cube roots continuously by using $\theta/3$. The reason $\theta$ can't be defined continuously is because if, say, the positive real axis is assigned $\theta = 0$, then by the time you wander all the way around a circle centered at 0, you find that the same point wants to be assigned either $\theta = 2\pi$ or $-2\pi$, depending on which direction you wander. This is explained in more detail here.
By taking a branch cut, we remove that curve from consideration and try to define $\theta$ continuously on the rest of the complex plane, and then divide it by 3 to take a cube root (i.e., define a branch of $g(z)$). This can be done. Suppose we take a branch cut going from 0 to $\infty$, upwards along the imaginary axis. We are given that Im[$g(1)$] = 0 (i.e. $g(1)=1$), and asked to find $g(-1)$. Since Im[$g(1)$] = 0, we should choose $\theta = 0$ along the positive real axis, so that $\theta/3 = 0$ agrees with the arg of the given value $g(1)=1$. Now we need to find a path from 1 to -1 which avoids the branch cut. This can be any curve joining 1 to -1 that swings downwards into the lower half plane. Because we swing downwards, by the time we get to -1, $\theta$ must be $-\pi$ (not $\pi$). Then $\theta/3$ = $-\pi/3$. So the cube root of -1 we want must have its argument equal to $-\pi/3$, so $g(-1) = e^{-\pi i/3}$.
Similarly, if we were to choose a branch cut going downwards from 0 to $\infty$, then $g(-1) = e^{\pi i/3}$ instead.
Now back to our original problem with $f(z)$. You choose a branch cut between $z_1$ and $z_2$, and find a path between 10 (where you are given a value) to 1 (where you need to find the value). This path must avoid the branch cut. Near each of the branch points, you have a similar situation as in the $g(z)$ example above. I'll let you work out the details.
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Thank you! So looking at the original problem I know Im[$f(10)$] = 0, so I know my initial location is 10 along the realy axis. This will give me an initial value of $\theta = 0$. To get from $f(10)$ to $f(1)$ I would need to travel around either the branch cut $z_1$ or $z_2$, thus traveling $\pi$ or $-\pi$. So, taking my original function, $f(z) = (\frac{8}{7}z^3-\frac{64}{7})^\frac{1}{3}$ I can rewrite this as $f(z) = (\frac87r^3\exp(3\theta i)-\frac{64}{7})^\frac{1}{3}$. Then plugging in $r=1$ and $\theta = \pi$ I get $f(1) = (\frac{8}{7}exp(3\pi i)-\frac{64}{7})^\frac{1}{3}$, right? – Jim Feb 14 '14 at 07:17
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This isn't right: (1) Your final answer cannot have a 1/3 exponent in it, or else you really haven't specified which branch you're taking. (It would be like writing $g(-1)=(-1)^{1/3}$ for the $g(z)$ example... which cube root of -1 do you mean? That's the whole point of the question.) (2) The branch point in the $g(z)$ example was at 0, but for $f(z)$, the branch points are elsewhere. You need to shift the coordinates to put a branch point (either $z_1$ or $z_2$) to 0 before you can apply the $g(z)$ example to this problem. – Ted Feb 14 '14 at 08:42
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So I'm going to try to do what you said in part (2). So to do this I am going to shift coordinates to make $z1$ at 0. So I would say $w=z-2$, rearranging $z=w+2$. Substituting this into f(z) I get $f(w) = (\frac87 (w+2)^3-\frac{64}{7})^\frac13$. This reduces down to $f(w) = (\frac87(w^3+6w^2+12w))^\frac13$ This has roots at w = 0, $2(1-exp(2\pi i/3))$, and $2(1-exp(-2\pi i/3))$. These are those of $f(z)$ but shifted to the left by 2. So for Im[f(10)]= 0 I know that $\theta = 0$. I can not see exactly I would go about getting rid of the 1/3 exponent. I've tried several things but can't get it. – Jim Feb 15 '14 at 00:35
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When I said that you can't have a 1/3 exponent, perhaps I should have said instead: you can use a 1/3 exponent, but you have to specify which cube root you mean. Anyway, in the new coordinates, you take a path joining w=8 to w=-1. Since your branch point is now at 0, you can factor out a $w^{1/3}$. That factor is exactly the $g(z)$ example. The remaining factor $w^2 + 6w + 12$ has no real roots between 8 and -1 (assuming you did the algebra correctly, I haven't checked) so you can take the straight path from 8 to -1, which (given the initial value at 8) is just taking the real cube root. – Ted Feb 15 '14 at 06:40
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I followed all of that up until you said we can take the straight path from 8 to -1. Wouldn't that make the path pass through the branch cut. I thought the whole purpose of the branch cut was that we couldn't go through them. Wouldn't you have to go around the branch cut? Thus making $\theta = \pi\ or -\pi$? – Jim Feb 16 '14 at 23:16
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For $(w^2+6w+12)^{1/3}$ there is no branch point at 0. For that factor, it doesn't matter whether you go through 0 or around it; it will lead to the same value. Only the $w^{1/3}$ factor is affected by the branch point at 0. For $(w^2+6w+12)^{1/3}$ it's a little easier to work out if you take a direct path, because then all the values of $w^2+6w+12$ are real and positive, so you can take the real cube root function as the branch. – Ted Feb 17 '14 at 00:45