I'm trying to evaluate the integral of $\frac{z^{\frac{1}{3}}}{(z+1)^2}$ on the following keyhole contour. enter image description here

I am using the branch cut of $[0, \infty)$.

In the solutions to this question, it is said that along the green line we can write $z=xe^{0 i}$ and along the red we can write it as $z=xe^{2 \pi i}$.

Can someone explain why this is so? I'm having a hard time understanding why, because the green and red line avoid the branch cut, and hence the arguments aren't actually $0, 2\pi$ respectively.

I'm also struggling to understand a branch cut for $z^a$ for $a$ not an integer. My understanding is that it is any line joining $0$ and '$\infty$', as these are the branch points. Could someone explain why these are the branch points?

  • $\begingroup$ Isn't it just the sum of the residues (times 2pi)? $\endgroup$ – mathworker21 Apr 21 '17 at 17:07
  • $\begingroup$ @mathworker21 Yes, but the question is in relation to finding the integral of $\frac{x^{\frac{1}{3}}}{(x+1)^2}$ from $0$ to $\infty$. So I'm using the method of finding the integral on the closed curve (as you said), then showing the line integrals on all but the red and green lines tend to zero as the inner circle's radius tends to zero and the outer's to infinity. $\endgroup$ – hhattiecc Apr 21 '17 at 17:11
  • 2
    $\begingroup$ in the limit where you get infinitly close to the branch cut with your contour the argument is indeed $0$ or $2\pi$ $\endgroup$ – tired Apr 21 '17 at 17:12


The complex logarithm function is a multi-valued function that is defined as

$$\log(z)=\log(|z|)+i\arg(z) \tag1$$

where $\arg(z)$ is the multivalued argument of $z$.

The function $f(z)=z^c$, where $c\in \mathbb{C}$, is defined as

$$f(z)=e^{c\log(z)} \tag2$$

Therefore, $f(z)$ is also multivalued when $c$ is not an integer.


If $z_0$ is branch point of the multivalued function $f(z)$ then there is no open neighborhood $N(z_0)$ of $z_0$ on which $f$ is continuous. Loosely speaking, we cannot encircle $z_0$ without encountering a discontinuity.

We can see from $(1)$ that $z_0=0$ is a branch point of $\log(z)$. Let $z_0=e^{i\theta_0}$ be a point on the unit circle. Then $\log(z_0)=i\theta_0$.

We travel on the unit circle from $z_0$ by increasing $\arg(z)$ from $\theta_0$ to $\theta_0+2\pi$. While we have returned to $z_0$, the value of $\log(z)$ has jumped from $i\theta_0$ to $i(\theta_0+2\pi)$. (Note that we have tacitly cut the plane along the ray $\theta=\theta_0$).

Inasmuch as $(2)$ defines $z^c$, then for non-integer $c$, $z^c$ shares the branch point singularity of $\log(z)$.

To see the reason that $z=\infty$ is also a branch point, we let $w=1/z$. Since $\log(w)$ has a branch point at $w=0$, then $\log(z)=\log(1/w)$ has a branch point at $\infty$.


From the previous discussion, we know that $z^{1/3}$ has logarithmic branch points at $z=0$ and $z=\infty$. We choose to cut the plane along the positive real axis.

With this choice of branch cut, if we approach a point on the positive real axis along a contour in the first quadrant, then $\arg(z)$ approaches $0$. If we approach a point on the positive real axis along a contour in the fourth quadrant, then $\arg(z)$ approaches $2\pi$.

Referring to the diagram in the OP, we can formally parameterize the green (red) segments as $z=x \pm i\epsilon$, $x\in [\sqrt{\nu^2-\epsilon^2},\sqrt{R^2-\epsilon^2})$, where $\nu>0$ is the radius of the blue-colored circular arc centered at the origin and $R$ is the radius of the gray-colored circular arc. Then, we have

\begin{align} \lim_{\epsilon\to 0}\int_{\sqrt{\nu^2-\epsilon^2}}^{\sqrt{R^2-\epsilon^2}}\frac{(x+ i\epsilon)^{1/3}}{(x+ i\epsilon+1)^2}\,dx&=\int_\nu^R \frac{x^{1/3}}{(x+1)^2}\,dx\\\\ \lim_{\epsilon\to 0}\int_{\sqrt{\nu^2-\epsilon^2}}^{\sqrt{R^2-\epsilon^2}}\frac{(x- i\epsilon)^{1/3}}{(x- i\epsilon+1)^2}\,dx&=\int_\nu^R \frac{x^{1/3}e^{i2\pi/3}}{(x+1)^2}\,dx \end{align}


It can be shown that as $\nu\to 0$ and $R\to \infty$, the contributions from the integrals around the circular arcs vanish. This leaves

$$\begin{align} (1-e^{i2\pi/3})\int_0^\infty \frac{x^{1/3}}{(x+1)^2}\,dx&=2\pi i \text{Res}\left(\frac{z^{1/3}}{(z+1)^2},z=-1\right)\\\\ &=2\pi i\lim_{z\to -1}\frac13z^{-2/3}\\\\ &=\frac{2\pi i}3 e^{-i2\pi/3} \end{align}$$

Solving for


  • $\begingroup$ Thanks for sharing the details on the last paragraph. I always assume the right hand side of the two last equations. Big (+1) . $\endgroup$ – Zaid Alyafeai Apr 22 '17 at 4:03
  • $\begingroup$ So you are using that $\nu^2 = \epsilon^2 + x_0^2 \to x_0 = \sqrt{\nu^2-\epsilon^2}$ , for the initial $x$ ? $\endgroup$ – Zaid Alyafeai Apr 22 '17 at 4:15
  • $\begingroup$ @ZaidAlyafeai You're welcome. And thank you as always! The initial value of $x$, prior to taking any limit, is $\sqrt{\nu^2-\epsilon^2}$, where we assume that $\epsilon<\nu$. $\endgroup$ – Mark Viola Apr 22 '17 at 4:17
  • $\begingroup$ Nice job. Nice answer! $\endgroup$ – PiE Apr 22 '17 at 4:26
  • $\begingroup$ Just edited your answer, LaTeX seemed to not work after your last edit. $\endgroup$ – Zaid Alyafeai Apr 22 '17 at 4:27

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