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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?

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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 $$f(z) = \cases{-\sqrt{z^2-1} & for $\text{Re}(z) \ge 0$\cr +\sqrt{z^2-1} & for $\text{Re}(z) < 0$\cr}$$

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

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