Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following question:

Show there is a single-valued analytic branch $f(z)$ for $\sqrt{z^2-1}$ in $\mathbb{C}\backslash [-1,1]$ such that $f(x) < 0$ for $x>1$. Here $[-1,1]$ denotes a closed interval in $\mathbb{R}$.

The following solution mimics a derivation from Moore and Hadlock's text.

Note that $z^2-1 = (z-1)(z+1)$. Consider the following four functions: \begin{align*} w_1(z) &= \sqrt{|z-1|} e^{(i\operatorname{arg_1}(z-1))/2} &\text{ where } \operatorname{arg_1}(z-1) \in [0,2\pi) \\ w_2(z) &= \sqrt{|z+1|} e^{(i\operatorname{arg_1}(z+1))/2} &\text{ where } \operatorname{arg_1}(z+1) \in [0, 2\pi) \\ w_3(z) &= \sqrt{|z-1|} e^{(i\operatorname{arg_2}(z-1))/2} &\text{ where } \operatorname{arg_2}(z-1) \in [-\pi,\pi) \\ w_4(z) &= \sqrt{|z+1|} e^{(i\operatorname{arg_2}(z+1))/2} &\text{ where } \operatorname{arg_2}(z+1) \in [-\pi, \pi) \end{align*} Note that

$w_1$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z-1) \neq 0\} = \mathbb{C} \backslash [1, \infty)$;

$w_2$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z+1) \neq 0\} = \mathbb{C} \backslash [-1, \infty)$;

$w_3$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z-1) \neq \pi\} = \mathbb{C} \backslash (-\infty, 1]$; and

$w_4$ is analytic on the set $\{ z \in \mathbb{C}: \operatorname{arg}(z+1) \neq \pi \} = \mathbb{C} \backslash (-\infty, -1]$.

Therefore $w_1w_2$ is analytic on the set $\mathbb{C} \backslash [-1, \infty)$ and $w_3w_4$ is analytic on the set $\mathbb{C} \backslash (-\infty, 1]$.

Furthermore, on the set $\mathbb{C} \backslash (-\infty, 1]$, we have $w_1w_2 = w_3w_4$, as we see below:

If $x > 1$, then $\operatorname{arg_1}(x \pm 1) = \operatorname{arg_2}(x \pm 1) = 0$, and thus $$w_1w_2(x)= w_3w_4(x) = \sqrt{x^2-1} \cdot e^{i \cdot 0} = \sqrt{x^2-1}. $$

If $\operatorname{Im}(z) > 0$, then $w_1w_2(z) = w_3w_4(z)$ since $\operatorname{arg}(z-1), \operatorname{arg}(z+1) \in (0, \pi)$.

If $\operatorname{Im}(z) < 0$, then $\operatorname{arg_1}(z\pm 1) = 2\pi + \operatorname{arg_2}(z\pm 1)$. Then $$w_1w_2(z) = \sqrt{|z^2-1|} e^{i(\operatorname{arg_1}(z-1) +\operatorname{arg_1}(z-1))/2} = \sqrt{|z^2-1|} e^{i(\operatorname{arg_2}(z-1) +\operatorname{arg_2}(z-1))/2 + 2\pi i } $$ $$ = \sqrt{|z^2-1|} e^{i(\operatorname{arg_2}(z-1) +\operatorname{arg_2}(z-1))/2} = w_3w_4(z).$$

This implies that $w_1w_2$ is analytic not only on $\mathbb{C} \backslash [-1, \infty)$, but on the larger set $\mathbb{C} \backslash [-1, 1]$. This is a single-valued analytic branch of $\sqrt{z^2-1}$. However, for $x >1$, we have $w_1w_2(x) >0$.

To address this, multiply $w_1w_2$ by $e^{i\pi}$. The resulting function is a single-valued analytic branch with the desired property.

I understand the computations here, but I'm trying to grasp the motivation behind the definitions, and see how it might generalize. Is there a way to attack general problems of this type, or a simple explanation of why these particular $w$ functions were chosen?

share|cite|improve this question
up vote 3 down vote accepted

EDIT: A more straightforward solution is to note that $z^2 - 1$ is in $(-\infty,0]$ exactly when $z$ is in the interval $[-1,1]$ or the imaginary axis. If we use the principal branch of $\sqrt{}$, $\sqrt{z^2-1}$ will be analytic everywhere else. This would be positive on $[1,\infty)$, so instead we take $-\sqrt{z^2-1}$. To avoid having a branch cut on the imaginary axis, we switch to $+\sqrt{z^2-1}$ in the left half plane. Thus $$f(z) = \cases{-\sqrt{z^2-1} & for $\text{Re}(z) \ge 0$\cr +\sqrt{z^2-1} & for $\text{Re}(z) < 0$\cr}$$

share|cite|improve this answer
But isn't $z^2 - 1 \in (- \infty, 0]$ for all $z$ on the imaginary axis also? – ec92 Jul 5 '12 at 19:40
Oops, yes: fixed it. – Robert Israel Jul 5 '12 at 19:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.