# Branch cut and contour integration along a special function

Define for all complex $z$ except for a slit on the real interval $[0,1]$, the analytic function $f(z)=(z^2-z^3)^{-1/3}$, so that $f(z)$ is real valued on the upper side of the slit.

a) How are the values of $f(z)$ on the lower side of the slit related to those on the upper side of the slit

b) Compute $\int_\gamma f$ taken once around any smooth curve $\gamma$ enclosing the slit.

For a), should I pick the branch cut on the interval $[0,1]$. Namely, $0<\arg z<2\pi$ and $-\pi <\arg (z-1)<\pi$. Then we see that the difference between the argument is $\pi/3$ between the top and bottom of the slit.

For b), the gamma is taken around the interval $[0,1]$, so we should just use Cauchy integral formula correct? But this is different from the answer given.

First note that $f(z)=(z^2-z^3)^{-1/3}$ has branch points at $z=0$ and $z=1$. Thus, we can choose to take as the branch cut the real-line segment $[0,1]$. If we do so, then we must still define how the arguments of both $z^{-2/3}$ and $(1-z)^{-1/3}$ are defined.

We can think of the branch cut (" the slit") as comprised of two branch cuts, one from $(0,0)$ to $(\infty,0)$ and the other from $(1,0)$ to $(\infty,0)$. Alternatively, we could take the branch cuts to be from $(0,0)$ to $(-\infty,0)$ and the other from $(1,0)$ to $(-\infty,0)$.

If we adopt the former convention for defining the arguments of both $z^{-2/3}$ and $(1-z)^{-1/3}$, then we have

$(i)$ On the "upper slit," the argument for $z$ is $0$ and the argument for $1-z$ is also $0$. Thus, the argument of $f(z)$ is $0$.

$(ii)$ On the "lower slit," the argument for $z$ is $2\pi$ while the argument for $1-z$ remains $0$. Thus, the argument for $f(z)$ is $-4\pi/3$.

If we adopt the latter convention for defining the arguments of both $z^{-2/3}$ and $(1-z)^{-1/3}$, then we have

$(i)$ On the "upper slit," the argument for $z$ is $0$ and the argument for $1-z$ is also $0$. Thus, the argument of $f(z)$ is $0$.

$(ii)$ On the "lower slit," the argument for $z$ is $0$ while the argument for $1-z$ is $-2\pi$. Thus, the argument for $f(z)$ is $2\pi/3$.

IMPROTANT NOTE:

It is important to note that $e^{i2\pi/3}=e^{-i4\pi/3}$ and that the results for the two alternatives explored here are identical.

METHODOLOGY 1: Direct Integration of $f$ around $\gamma$

The integral $I$ around the contour $\gamma$ of the slit can be written

\begin{align} I&=\oint_{\gamma}(z^2-z^3)^{-1/3}dz\\\\ &=\int_0^1 x^{-2/3}(1-x)^{-1/3}dx+\int_1^0e^{i2\pi/3}x^{-2/3}(1-x)^{-1/3}dx\\\\ &=\left(1-e^{i2\pi/3}\right)\int_0^1x^{-2/3}(1-x)^{-1/3}dx\\\\ &=\left(1-e^{i2\pi/3}\right)\text{B}\left(\frac13,\frac23\right) \tag 1\\\\ &=\left(1-e^{i2\pi/3}\right)\frac{\Gamma\left(\frac13\right)\Gamma\left(\frac23\right)}{\Gamma\left(\frac13+\frac23\right)} \tag 2\\\\ &=\left(1-e^{i2\pi/3}\right)\frac{\Gamma\left(\frac13\right)\Gamma\left(1-\frac13\right)}{\Gamma\left(1\right)} \\\\ &=\left(1-e^{i2\pi/3}\right)\frac{\pi}{\sin(\pi/3)} \tag 3\\\\ &=\left(1-e^{i2\pi/3}\right)\frac{2\pi}{\sqrt{3}}\\\\ &=\pi(\sqrt{3}+i) \end{align}

Thus, we have

$$\bbox[5px,border:2px solid #C0A000]{I=\oint_{\gamma}(z^2-z^3)^{-1/3}dz=\pi(\sqrt{3}+i)}$$

Special Notes: In $(1)$, $\text{B}(x,y)$ is the Beta Function.

In $(2)$, $\Gamma(x)$ is the Gamma Function, which is related to the Beta Function by $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

In $(3)$, we used Euler's Reflection Principle, $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin (\pi x)}$.

METHODOLOGY 2: Using the Residue at Infinity to Integrate of $f$ around $\gamma$

Another way to evaluate the integral $I$ is to evaluate the Residue at Infinity. The residue at infinity is given by

\begin{align} \text{Res}\left(f,\infty \right) &= \text{Res}\left(-\frac{1}{z^2}f\left(\frac{1}{z}\right),z=0\right)\\\\ &=\lim_{z\to 0}\,z\,\left(-\frac{1}{z^2}\left(\frac{z-1}{z^3}\right)^{-1/3}\right)\\\\ &=-e^{-i\pi/3} \end{align}

Thus,

$$\bbox[5px,border:2px solid #C0A000]{I=-2\pi i (-e^{-\pi/3})=\pi(\sqrt{3}+i)}$$

as expected!

• One more question. Why would you take the branch cut both on the real axis? Now the branch cut is on $[0,\infty)$. Shouldn't you put your slit on $[0,1]$
– nerd
Aug 6, 2015 at 2:07
• The answer is that we do take the branch cut as $[0,1]$. This slit can be decomposed as two branch cuts, each of which begins at a branch point and extends, for example, along the positive real axis to $\infty$. But, the portion for which the two cuts overlap (i.e., from $1$ to $\infty$) "cancel" in the sense that $f$ does not change value upon crossing the line at points for which $x>1$. Thus, $[0,1]$ is the effective branch cut. Please let me know if this helps and how I can further improve my answer. I really want to give you the best one I can. Aug 6, 2015 at 2:52
• I see, so whenever the branch collide, it cancel the effect: For $\arg z^{-2/3}$ the argument below the slit is $-4\pi/3$, and for $\arg(1-z)^{-1/3}$ is $-2\pi/3$, so that gives $-2\pi\equiv 0$. Correct?
– nerd
Aug 6, 2015 at 5:50
• @nerd Yes. You have it now. Well done! Aug 6, 2015 at 13:50
• @nerd Just a word of caution .... the "cancellation effect" of the two branch cuts is present here, but is not true in general. If we replace $f$ with $(z(1-z))^{-1/3}$ then the cuts do not cancel and thus the slit is not appropriate. If we replace $f$ with $(z(1-z))^{-1/2}$, then we observe that the "cancellation effect" works and a slit here would work too. So, it is a case by case basis ... we just need to check each time. Aug 6, 2015 at 14:08