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I'm working on a problem in Gamelin's Complex Analysis (Chapter IV, Section 2, page 109, exercise #4). I'm asked to consider the branch of $f(z)=\sqrt{z^2-1}$ on $D=C\setminus (-\infty,1]$ that is positive on the interval $(1,\infty)$.

Under these conditions, one can supposedly show that $F(z)=\log\left(z+\sqrt{z^2-1}\right)$ is analytic on $D$ and $F'(z)=f(z)$. Thus, if $\gamma(t)=2e^{i\theta}$, where $-\pi/2\le\theta\le\pi/2$, then:

$$\begin{align*} \int_{\gamma}\frac{dz}{\sqrt{z^2-1}} &=\log\left(z+\sqrt{z^2-1}\right)\bigg|_{-2i}^{2i}\\ &=\log\left(2i+\sqrt{(2i)^2-1}\right)-\log\left(-2i+\sqrt{(-2i)^2-1}\right)\\ &=\log\left(2i+\sqrt{-5}\right)-\log\left(-2i+\sqrt{-5}\right)\\ &=\log(2i+\sqrt5i)-\log(-2i+\sqrt5i) \end{align*}$$

The last step here is very suspicious to me at the moment. Continuing:

$$\begin{align*} \int_{\gamma}\frac{dz}{\sqrt{z^2-1}} &=\log((2+\sqrt5)i)-\log((-2+\sqrt5)i)\\ &=\left[\log(2+\sqrt5)+i\frac{\pi}{2}\right]-\left[\log(-2+\sqrt5)+i\frac{\pi}{2}\right]\\ &=\log\frac{2+\sqrt5}{-2+\sqrt5}\\ &=\log(9+4\sqrt5) \end{align*}$$

Mathematica agrees, although I think they have a different definition for the branch cut.

g = Function[z, Log[z + Sqrt[z^2 - 1]]]

g[2 I] - g[-2 I] // ComplexExpand

Output:

log(2 + Sqrt[5]) - log(Sqrt[5] - 2)

Now, Murray Eisenberg wrote some nice code to compute a contour integral using midpoint sums.

enter image description here

Now, when I take advantage of this routine, here is the result.

enter image description here

Which of course, suggests that the answer is $i\pi$.

Any thoughts?

Additional Comment I've come up with this image:

enter image description here

For any $z$ not on the cut, we have

$$ \begin{align*} z-1&=r_1e^{i(\theta_1+2k\pi)}\\ z+1&=r_2e^{i(\theta_2+2m\pi)}\\ \end{align*} $$

where $r_1$ and $r_2$ are the lengths (magnitudes) of $z-1$ and $z+1$. Hence: $$ \begin{align*} \sqrt{z^2-1} &=\sqrt{z-1}{z+2}\\ &=\sqrt{r_1}\sqrt{r_2}e^{i(\theta_1/2+\theta_2/2+(k+m)\pi)} \end{align*} $$

When we evaluate this expression on the real axis and get a positive result for $x>1$, this tells us that $k+m$ must be an even integer. When $k+m$ is odd, the sign of the answer changes. Hence, let's use $k+m=0$ for our branch and write:

$$\sqrt{z^2-1}=\sqrt{r_1}\sqrt{r_2}e^{i(\theta_1/2+\theta_2/2)}$$

Now, when $z=2i$, we can show that $$ \begin{align*} \theta_1&=\frac{\pi}{2}-\tan^{-1}2\\ \theta_2&=\tan^{-1}2, \end{align*}$$ so $\theta_1/2+\theta_2/2=\pi/2$. Hence, at $z=2i$, $\sqrt{z^2-1}$ will equal its magnitude times $e^{i\pi/2}$, that is, times $i$.

On the other hand, note that $\alpha_1/2+\alpha_2/2=-\pi/2$, so at $z=-2i$, $\sqrt{z^2-1}$ will have the same magnitude, but times $e^{-i\pi/2}$, or times $-i$.

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  • $\begingroup$ There's also the issue of what branch of log is being used in your symbolic evaluation. $\endgroup$
    – murray
    Mar 12, 2013 at 22:38
  • $\begingroup$ Gamelin has you show that $z+\sqrt{z^2-1}$ is never nonpositive and real on $D$ in part (a) of this exercise, so you can use the principal branch of the logarithm. $\endgroup$ Mar 12, 2013 at 22:41

1 Answer 1

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If $\phi(z)$ is the branch of $\sqrt{z^2-1}$ with a branch cut at $(-\infty,1]$ which is positive on $(1,\infty)$, then $\phi(2i)=\sqrt{5}i$ and $\phi(-2i)=-\sqrt{5}i$. So (writing $\text{Log}$ for the principal value of the logarithm as Gamelin does) $$\begin{align*} \text{Log}\left(z+\phi(z)\right)\bigg|_{-2i}^{2i} &=\text{Log}\left(2i+\phi(2i)\right)-\text{Log}\left(-2i+\phi(-2i)\right)\\ &=\text{Log}((2+\sqrt{5})i)-\text{Log}((2+\sqrt{5})(-i))\\ &=\pi i, \end{align*}$$ in agreement with the numerical result.

To see that $\phi(2i)=\sqrt{5}i$, let $z$ move along a path from $2$ to $2i$ in the closed first quadrant, for example, $\theta\mapsto z(\theta):= 2e^{i\theta}$, $0\le \theta\le \pi/2$. Then $z(\theta)^2-1=4 e^{2i\theta}-1$ will travel from $3$ to $-5$, always remaining in the closed upper half-plane. Since $\phi(z)$ was defined to be positive on $(1,\infty)$, $\phi(2)=\sqrt{3}>0$, so $\phi(z)$ must also travel along a path in the closed first quadrant. Therefore $\phi(2i)$ is $\sqrt{5}i$ rather than $-\sqrt{5}i$. Proving that $\phi(-2i)=-\sqrt{5}i$ can be done in the same way, by moving $z$ on a path from $2$ to $-2i$ in the closed fourth quadrant and observing that, since $z^2-1$ moves along a path in the closed lower half-plane, $\phi(z)$ must travel along a path in the closed fourth quadrant.

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  • $\begingroup$ I think I agree. Now, the issue here is "Why is $\phi(-2i)=-\sqrt5i$? How can we show this? $\endgroup$
    – David
    Mar 12, 2013 at 23:01
  • $\begingroup$ I commented on this above. $\endgroup$ Mar 12, 2013 at 23:12

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