# Riemann surface of $f(z)=((z-1)(z-2)(z-3))^{2/3}$

I try to describe the Riemann surface of $f(z)=((z-1)(z-2)(z-3))^{2/3}$. I found the branch points 1,2, and 3 also realized $\infty$ is not a branch point. Since we take third root, I see three sheet. I am not sure but the line segment $[1,3]$ is my candidate for branch cut. Because in this case i prevent to go around the points 1,2,3. My questions are the followings:

• What is the appropriate branch cuts in this question?
• Is its Riemann surface homeomorphic to any well-known surface?
• Do different branch cuts give rise to same Riemann surface as a topological space?
• Are branch points among zeroes and poles

Also I would be grateful if you could recommend some references about these questions. Thanks

Write it as $w^3-(z-1)^2(z-2)^2(z-3)^2=0$. The branch table is: $$\left( \begin{array}{cccc} 1 & 1.0000000 & \left( \begin{array}{ccc} 1 & 2 & 3 \\ \end{array} \right) & \\ 2 & 2.0000000 & \left( \begin{array}{ccc} 1 & 2 & 3 \\ \end{array} \right) & \\ 3 & 3.0000000 & \left( \begin{array}{ccc} 1 & 2 & 3 \\ \end{array} \right) & \\ 4 & \infty & \left( \begin{array}{c} 1 \\ 2 \\ 3 \\ \end{array} \right) & \\ \end{array} \right)$$

Thus by Riemann-Hurwitz, we have $1-3+1/2(2+2+2)=1$, a genus-1 surface or torus. An algebraic pole for the function: $$F(z,w)=a_1(z)+a_2(z)w+\cdots+a_n(z)w^n=0$$ is given by $a_n(z)=0$ so that in the case above, the function has no poles.

Use the logarithmic derivative to define the function: $$f = C\exp\left\{\int_{z_0}^{z}\frac{2}{3}\left[\frac{1}{w-1}+\frac{1}{w-2}+\frac{1}{w-3}\right]dw\right\}$$ If you choose a path from $z_0$ to $z$ that circles 3 times around one pole, 2 times around one pole and 1 time around another, or circles all three, the function returns to the original value. You can define a branch by omitting the segment from $1$ to $3$. Or, you can omit non-intersecting infinite segments from $1$ to $\infty$, from $2$ to $\infty$, and from $3$ to $\infty$.

We can choose any branch cuts that begin at the branch points $z=1,2,3$ and extend to the point at $\infty$.

CHOICE $1$:

As a first example, we cut the plane with the three straight line paths along the positive real axis starting at (i) $z=1$, (ii) $z=2$ and (iii) $z=3$ and ending at infinity.

Now, let's examine the argument of $f(z)$ when $\text{Im}(z)=0$ and $\text{Re}(z)>3$ on the sheet for which the $0\le \arg(z-n) <2\pi$ for $n=1,2,3$.

Approaching the real axis from the upper-half plane, we have $\arg(f(z))=0$ while approaching from the lower-half plane, we have $\arg(f(z))=\frac23(2\pi +2\pi +2\pi)=4\pi$. Inasmuch as $e^0 =e^{i4\pi}$, we see that $f(z)$ is single-valued on that part of the real axis where the three branch cuts coincide.

Therefore, the "effective" branch cut is simply the line segment from $z=1$ to $z=3$.

CHOICE $2$:

As a second example, we cut the plane with three straight line paths (i) along the real axis starting at (i) $z=1$ and ending at $z=-\infty$, and along the positive real axis starting at (ii) $z=2$ and (iii) $z=3$ and ending at infinity.

Let's examine the argument of $f(z)$ when $\text{Im}(z)=0$ and $\text{Re}(z)>3$ on the sheet for which $-\pi <\arg(z-1)\le \pi$ and $0\le \arg(z-n) <2\pi$ for $n=2,3$.

Approaching the real axis from the upper-half plane, we have $\arg(f(z))=0$ while approaching from the lower-half plane, we have $\arg(f(z))=\frac23(2\pi +2\pi +0)=\frac{8\pi}{3}$. Inasmuch as $e^0\ne e^{i8\pi/3}$, we see that $f(z)$ is not single-valued on that part of the real axis where only two of the three branch cuts coincide.

Next, let's examine the argument of $f(z)$ when $\text{Im}(z)=0$ and $\text{Re}(z)<1$ on the same sheet for which $-\pi <\arg(z-1)\le \pi$ and $0\le \arg(z-n) <2\pi$ for $n=2,3$.

Approaching the real axis from the upper-half plane, we have $\arg(f(z))=\frac23(\pi+\pi+\pi)=2\pi$ while approaching from the lower-half plane, we have $\arg(f(z))=\frac23(\pi +\pi -\pi)=\frac{4\pi}{3}$. Inasmuch as $e^{2\pi}\ne e^{i4\pi/3}$, we see that $f(z)$ is not single-valued on that part of the real axis where there is only one branch cut.

Therefore, $f(z)$ is analytic neither along the positive real axis from $z=2$ to $\infty$ nor along the real axis from $z=1$ to $z=-\infty$.