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I am still not convinced by the post that the function$$\sqrt[3]{(z-1)(z-2)(z-3)}$$ can be defined so it is analytic on $\mathbb{C}\backslash [1,3]$. We define for each $z\in \mathbb{C}\backslash (-\infty,3]$ the function $$f(z)=\int_4^z \frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}\,dz +\ln 6$$ and it claims that function $$\exp\left(\frac{1}{3}f(z)\right)$$ is continuous on $(-\infty,1)$. This is the part I don't understand. How do we prove that it is continuous on $(-\infty,1)$. I computed the integral from 4 to $-4$ along the upper semicircle of $|z|=4$, and the integral from 4 to $-4$ along the lower semicircle of $|z|=4$ and they are not equal. So I don't see how the function can be continuous at $-4$, for example.

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TCL
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  • My computation was done by Maple, and obviously, it made a mistake. – TCL May 14 '13 at 12:32
  • This first integral should be $\ln 35 +3\pi I$ and the second $\ln 35 - 3\pi I$. Maple gives $\ln 35+\pi I $ and $\ln 35 -\pi I$. Now I understand. – TCL May 14 '13 at 12:47

2 Answers2

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Given that $$\frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}=\frac1{z-1}+\frac1{z-2}+\frac1{z-3}$$ you see that all the three residues of the integrand are equal to one, so their sum is $3$. The "upper" and "lower" integrals thus differ by $6\pi i$.

After multiplication by $1/3$ they differ by $2\pi i$, and after $\exp$ they agree. IOW $f(z)$ won't be continuous, but $\exp(f(z)/3)$ still is.

Jyrki Lahtonen
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    Note that it is essential that the number of factors was a multiple of three. For example, you won't get a cube root of $(z-1)(z-2)(z-3)(z-4)$ continuous on the complement of $[1,4]$ nor a continuout cube root $(z-1)(z-2)$ on the complement of $[1,2]$ by analogous formulas. – Jyrki Lahtonen May 14 '13 at 06:06
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If you believe in the Monodromy theorem and are comfortable with the Riemann sphere $\overline{\mathbb C}$, the following short argument suffices: $$\Big((z-1)(z-2)(z-3)\Big)^{1/3} = z\Big((1-1/z)(1-2/z)(1-3/z) \Big)^{1/3}$$ Here $\Big((1-1/z)(1-2/z)(1-3/z) \Big)^{1/3}$ admits analytic continuation in $\overline{\mathbb C}\setminus [1,3]$, which is a simply-connected domain. Therefore, it has a single-valued branch there.

If you are not comfortable with the Riemann sphere, but still believe in the Monodromy theorem, then move the segment to infinity, for example by letting $w=1/(z-1)$. This transforms $\overline{\mathbb C}\setminus [1,3]$ into $ {\mathbb C}\setminus [1/2,\infty)$. We get $$\Big((z-1)(z-2)(z-3)\Big)^{1/3} = \Big( w^{-1}(w^{-1}-1) (w^{-1}-2) \Big)^{1/3} =w^{-1} ( (1-w)(1-2w) )^{1/3} $$ Here $( (1-w)(1-2w) )^{1/3}$ admits analytic continuation in $ {\mathbb C}\setminus [1/2,\infty)$, which is a simply-connected domain. Therefore, it has a single-valued branch there.

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