Take the 2-minute tour ×
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.

From Stein/Shakarchi's Complex Analysis page 232:

...We consider for $z\in \mathbb{H}$, $$f(z)=\int_0^z \frac{d\zeta}{(1-\zeta^2)^{1/2}},$$ where the integral is taken from $0$ to $z$ along any path in the closed upper half-plane. We choose the branch for $(1-\zeta^2)^{1/2}$ that makes it holomorphic in the upper half-plane and positive when $-1<\zeta<1$. As a result, $$(1-\zeta^2)^{-1/2}=i(\zeta^2-1)^{-1/2}\quad \text{when }\zeta>1.$$

I'm a little confused on the branch they're choosing. Do they mean that the we cut along $\mathbb{R}\setminus [-1,1]$? But then $(\zeta^2-1)^{-1/2}$ is undefined for $\zeta>1$...

share|improve this question
Since the whole question plays in the upper half plane, the cut is along the full x axis. To get to the mentioned real points, cut along some rays extending into the negative half plane like $\pm1-i\mathbb R_+$. –  LutzL Jan 3 at 20:01

2 Answers 2

up vote 6 down vote accepted

A good way to think about powers is through the logarithm $$ (1-\zeta^{2})^{1/2} = e^{\frac{1}{2}\log(1-\zeta^{2})}. $$ Wherever the logarithm is holomorphic, so is $(1-\zeta^{2})^{-1/2}$. Wherever the logarithm is real, the square root is real and positive. Logarithm of $f(z)=(1-z^{2})$ is done using an integral of $f'(z)/f(z)$. In this case, $$ \log(1-\zeta^{2})=\int_{0}^{\zeta}\frac{-2z}{1-z^{2}}\,dz. $$ The only trick in designing the logarithm is to restrict the domain so that you cannot circle a point where there is a non-zero residue. So the definition of the above works just fine, so long as you choose any path from 0 to $\zeta$ along a path which does cross the real axis on $(-\infty,-1]$ or on $[1,\infty)$. That's how branch cuts are handled. Branch cuts can be along just about any curve you like, chosen to make sure you cannot circle the places where the integrand has non-zero residues. Then $$ (1-\zeta^{2})^{1/2} = \exp\left\{-\frac{1}{2}\int_{0}^{\zeta}\frac{2z}{1-z^{2}}dz\right\}=\exp\left\{\frac{1}{2}\int_{0}^{\zeta}\frac{1}{z-1}+\frac{1}{z+1}\,dz\right\}. $$ You can see that this function is positive and real on the real axis $(-1,1)$ because the integral reduces to one on the real line where the integrand is real. Furthermore, function is discontinuous across the real axis on $(-\infty,1]$ by a multiplicative constant equal to $\exp\{\frac{1}{2}(-2\pi i)\}=-1$ because of the residue of $\frac{1}{1+z}$ at $z=-1$.

share|improve this answer
Thank you! In the second paragraph, did you mean "...cross the real axis on $(-\infty, -1)$ or on..."? You wrote "$1$". (A similar thing is in the last paragraph). –  Eric Auld Jan 3 at 20:54
@Eric Auld: Thank for for that correction. I fixed that typo in the answer now. –  T.A.E. Jan 3 at 20:57
So then how are they defining the function at $\zeta>1$? These points are removed from the domain, right? Or it is defined there, but discontinuous, perhaps? –  Eric Auld Jan 3 at 20:59
@Eric Auld: If you choose a branch cut of a segment between the two points $z_1$ and $z_2$, then you you won't have a problem because any closed path integral of $1/(z-z_1)+1/(z-z_2)$ is either 0 or an integer multiple of $4\pi i$. So the square root still works in that case as $e^{1/2(4\pi i)}=1$. This is an extension of the result to specific powers. –  T.A.E. Jan 3 at 21:36
@Eric Auld: I forgot to mention: think logarithmic derivative $f'(z)/f(z)$. So you get $1/(z-z_1)+1/(z-z_2)$ for the integrand. –  T.A.E. Jan 3 at 21:50

Another (probably not as good) way to look at it is to think $$ \sqrt{1-\zeta^2} = i \sqrt{\zeta -1} \sqrt{\zeta + 1} $$ Choose the (possibly different) branches for each of the square roots on the right side to give you what you need. It might look at first like you will have problems on $z \in [-1,1]$, but the discontinuities cancel each other.

For example, use the principle branch of $\sqrt{w}$ on $\zeta - 1$, but for $\zeta + 1$ use the branch of $\sqrt{w}$ where $-2\pi < \arg(w) < 0$.

Use $-2 \pi \le \arg(\zeta+1) < 0$ and $-\pi < \arg(\zeta -1) \le \pi$.
Then $-\pi \le \arg(\sqrt{\zeta+1}) < 0$ and $-\pi/2 < \arg(\sqrt{\zeta -1}) \le \pi/2$.

So on $-1 < z < 1$ you get $$ \arg(i \sqrt{\zeta -1} \sqrt{\zeta + 1} ) = \pi/2 + (-\pi) + \pi/2 = 0 $$ On $1 < z$ you get $$ \arg(i \sqrt{\zeta -1} \sqrt{\zeta + 1} ) = \pi/2 + (-\pi) + 0 = -\pi/2 $$ so $$ \arg(\frac{1}{i \sqrt{\zeta -1} \sqrt{\zeta + 1}} ) = \pi/2 = \arg(\frac{i}{\sqrt{t^2-1}}) $$ where $t = |\zeta| > 1$ and $\sqrt{t^2-1}$ uses the regular square root on $\mathbb{R}^+$.

share|improve this answer
Can you explain why the factor is $i$? It seems to me like the argument of $(1-\zeta)(1+\zeta)$ should differ by $2\pi$ between $\zeta<1$ and $\zeta>1$. Then taking half of that, and applying the exponential would give us a factor of $-1$ between the two locations. –  Eric Auld Jan 8 at 1:52
@EricAuld: Note that I'm taking the root of $\zeta - 1$ instead of $1-\zeta$. Just easier for me to think of it that way. That's where the factor of $i$ comes from. –  bryanj Jan 8 at 1:57
Hmm...sorry, I'm still not understanding. In the OP it refers to $$(1-\zeta^2)^{-1/2}=i(\zeta^2-1)^{-1/2}$$ for $\zeta>1$. If we look at the argument of $(1-\zeta)(1+\zeta)$, it seems like it should change by $\pi$ when we go by $\zeta=1$, and change again by $\pi$ as we go by $\zeta=-1$. Therefore it is changing by $2\pi$ total...halve that and you get $\pi$, apply the exponential and you get a factor of $-1$. So why is the factor $i$? –  Eric Auld Jan 8 at 2:00
See the edit. The point is that in this method (which isn't as nice as the method in the other answer), you think of $\sqrt{\zeta^2-1}$ as the product of two functions, and use a different branch of the square root on each linear factor. Hope that helps. –  bryanj Jan 8 at 4:46

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.