Consider the following function $$ w(z) = \sqrt{\frac{z-a}{z-b}} \,\,\, a,b \in \mathbb{R} $$ assume $a<b$. Utilizing the bipolar coordinates $ z - a = r_1 e^{i \theta_1}$ and $z-b =r_2 e^{i \theta_2} $ (where $\theta_1 \in [-\pi, \pi)$ and $\theta_2 \in [0, 2\pi)$ ). I was able to conclude that the function is continuous on the domain $(b, \infty)$. My reasoning: $$w(z) = \sqrt{r_1/r_2} \exp{\big[\frac{i(\theta_1 - \theta_2)}{2}\big]} = Re^{i\Theta}$$ $w$ 'jumps' whenever $\Theta$ 'jumps', because $R$ is fixed. These jumps in $\Theta$ occur whenever you cross the $x$-axis on $(-\infty, b)$. Beyond that, crossing the $x$-axis is fine since $\Theta$ changes continuously.

2 questions:

  1. Do you think my reasoning is correct?
  2. I heard that branch cuts connect branch points. Does this imply that $a$ is not a branch point?

EDIT: I don't have a choice in my coordinate system, I can't let both $\theta_1, \theta_2$ be between $0$ and $2 \pi$.


You seem to have made a mistake somewhere. The branch cut of $w$ connects $z=a$ and $z=b$ through the point at infinity. Consider $z < a$. Then $\theta_2 = \pi$, and $\theta_1$ can be $-\pi$ or $\pi$, illustrating one of the discontinuities across the branch cut of $w(z)$: $$\sqrt{\frac{r_1}{r_2}}e^{\frac{i(-\pi-\pi)}{2}}\neq \sqrt{\frac{r_1}{r_2}}e^{\frac{i(\pi-\pi)}{2}}$$ Similarly, consider $z>b$. Then $\theta_2$ can be $0$ or $2\pi$, and $\theta_1 = 0$, illustrating again, the discontinuity across the branch cut: $$\sqrt{\frac{r_1}{r_2}}e^{\frac{i(0-0)}{2}}\neq \sqrt{\frac{r_1}{r_2}}e^{\frac{i(0-2\pi)}{2}}$$ There is no discontinuity across $(a,b)$, since on that interval, $\theta_1 = 0$ and $\theta_2 = \pi$.

  • $\begingroup$ Yup, you're 100% right. I worked on it some more and found my mistake. As you said the function is continuous on $(a,b)$. Is it then correct to say that the branch cut of the region is: $\text{Re} z, z \in (\infty, a) \cup (b, \infty)$. $\endgroup$
    – talrefae
    Feb 15 '18 at 22:39
  • 1
    $\begingroup$ @talrefae Minor correction to what I said earlier. You can say the branch cut is on $(-\infty, a]\cup [b,\infty)$ for $a,b\in \Bbb R$. $\endgroup$
    – bames
    Feb 16 '18 at 8:01
  • $\begingroup$ That helps. Thanks! $\endgroup$
    – talrefae
    Feb 17 '18 at 0:12

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