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My professor gave me the above problem as test preparation, and I am feeling a bit lost at how to solve it. I looked up the answer in the back of my book, and I didn't really understand the logic behind it. I was hoping someone could shed some insight into the problem solving process since I will have a similar, but not identical, question on my test. Without further ado, here is the solution:

Let $$ f(z) = \sqrt[3]{(z-1)(z-2)(z-3)} = \text{exp}(\frac{1}{3}\text{log}[(z-1)(z-2)(z-3)]) $$

where

$$ \text{log}((z-1)(z-2)(z-3)) = \int_4^z{\frac{[(z-1)(z-2)(z-3)]'}{(z-1)(z-2)(z-3)}dz} + \text{log} (6) \text{, for } z \in \mathbb{C} \backslash (-\infty, 3]$$

Then, it can be shown that $ f(z) $ defines a function that is continuous at all points $ x \in \mathbb{R} \text{ such that } x \lt 1 $.

There are many things I don't understand about this, mostly related to the choice of logarithm. Why is the integral from 4 to z? Why do you add log(6)? Why is it defined on $ \mathbb{C} \backslash (-\infty, 3] $? How can I figure out how to solve problems like this in similar cases?

Furthermore, what is the easiest way to show continuity on the real axis where x < 1? Thanks for the help.

Max
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  • You could also argue that the only branch points are $z=1, z=2, $ and $z=3$. That is, argue that there is no branch point at infinity. – Random Variable May 12 '13 at 02:36
  • @Max: Can you tell me which book you took this problem from? I am still not convince about the continuity of the function on $(-\infty,1)$. Thank you. – TCL May 14 '13 at 04:47
  • It is from Bak and Newman's book titled Complex Analysis. Chapter 10 Exercise 17. I convinced myself using the Argument principle that any path you took from 4 to z would only depend on the endpoints. – Max May 14 '13 at 10:40

2 Answers2

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The constant 4 is an arbitrary choice on $\mathbb{C}\backslash(-\infty,3]$ and $\log 6$ is to match $f(4)=\sqrt[3]{6}$. Your integrand function is holomorphic on $\mathbb{C}\backslash(-\infty,3]$, so the integration is well defined $\mathbb{C}\backslash(-\infty,3]$. To show $x<1$, $f(x)$ is continuous, you need to show that integration does not depend on path up to $6\pi i$, hence $f(x)$ is well-defined on $(-\infty, 1]$.

Ma Ming
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  • Thanks this makes perfect sense. – Max May 12 '13 at 02:21
  • @Ma Ming: I don't understand this. I evaluate the integral over the path $z=4e^{it},0\le t\le \pi$, I get $ln5+ln7+\pi i$, and when I evaluate the integral over the negative of the path $z=4e^{it},-\pi\le t\le 0$, I get $ln 5+\ln 7-\pi i$. (The integrals were done on Maple.) If the function is continuous at $-4$, shouldn't these two integrals be the same? – TCL May 14 '13 at 02:23
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    @TCL There must be a mistake. $\int_{|z|=4}( \frac{1}{z-1}+\frac{1}{z-2}+\frac{1}{z-3})dz=3\times 2\pi i$. – Ma Ming May 14 '13 at 07:04
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Note that, the function has three branch points, namely $z=1,2,3$. So, in order to make the function a single-valued function you need to take a branch cut in the complex plane. In this case, we can take the interval $[1,3]$ as branch cut. This will make the function defined on $\mathbb{C}\backslash[1,3]$.

Mhenni Benghorbal
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