Consider integrating $f(z)=\sqrt{1-z^2}$ with a branch cut of $[-1,1]$ around the following contour.

$\gamma_1:[-1,1]\to\mathbb{C}, t\mapsto t+\epsilon i$

$\gamma_2:[-\pi/2,\pi/2]\to\mathbb{C}, t \mapsto 1+\epsilon e^{-it}$

$\gamma_3:[-1,1]\to\mathbb{C}, t\mapsto -t-\epsilon i$

$\gamma_4:[-3\pi/2,-\pi/2]\to\mathbb{C}, t \mapsto -1+\epsilon e^{-it}$

As $\epsilon$ tends to $0$ we have that $\int_{\gamma_2}f(z)dz$ and $\int_{\gamma_4}f(z)dz$ tend to $0$.

Now, $\int_{\gamma_1}f(z)dz$ tends to $\int_{-1}^{1}\sqrt{1-x^2}dx:=I$ and also $\int_{\gamma_3}f(z)dz$ also tends to $I$ (one minus sign from being on the other side of the branch cut, another one from reversing the lower/upper limits).

Now $I=\pi/2$

So the integral around the closed contour $\lim_{\epsilon\to 0}\int_{\gamma_1+\gamma_2+\gamma_3+\gamma_4} f(z) dz=\pi$

Note that $f$ is analytic in $\mathbb{C}\setminus[-1,1]$ with our choice of branch cut and so we can consider the contour integral around $\gamma=\gamma_1+\gamma_2+\gamma_3+\gamma_4$ as a closed contour around infinity.

i.e. $\int_{\gamma}f(z)dz=-2\pi i Res[f,z=\infty]$ ($-2\pi i$ instead of $2\pi i $ because we are going clockwise around infinity.)

Now, $Res[f,z=\infty]=Res[\sqrt{1-1/z^2},z=0]=\lim_{z\to 0}z\sqrt{1-1/z^2}=\lim_{z\to 0}\sqrt{z^2-1}=i$

So $\int_{\gamma}f(z)dz=-2\pi i (i)=2\pi$

And the two results do not agree. I am not confident my arguments about the pole at infinity. What exactly did I do wrong there?

  • $\begingroup$ You messed with the branches when going from $\sqrt{1-1/z^2}$ to $\sqrt{z^2-1}$, or you didn't define the residue at $\infty$ correctly. For $f$ is analytic on $|z| > 1$ then $\int_{|z|=2} f(z)dz = -\int_{|s|=1/2} \frac{f(1/s)}{-s^2}ds = 2i \pi Res(\frac{f(1/s)}{s^2},0)$ $\endgroup$
    – reuns
    Commented Aug 8, 2019 at 22:23

1 Answer 1


See this answer. You have implicitly used the condition that $f(x + i0) > 0$ for $-1 < x <1$ (otherwise you would get $I = -\pi$). With this condition, $f$ can be written as $$f(z) = -i z \sqrt {1 - \frac 1 {z^2}},$$ where $\sqrt z$ is the principal value of the square root.

$\gamma$ goes around the origin clockwise, therefore $I = +2 \pi i \operatorname{Res}_{z = \infty} f(z)$. By the binomial theorem, the Laurent expansion of $\sqrt {1 - 1/z^2}$ around infinity is $1 - 1/(2 z^2) + O(1/z^3)$, which gives $\operatorname{Res}_{z = \infty} f(z) = 1/(2 i)$.

  • $\begingroup$ I understand that $\sqrt{1-1/z^2}=1-1/(2z^2)+O(1/z^4)$. So, $f(z)=-iz+i/(2z)+O(1/z^3)$ and thus, $f(1/z)=-i/z+O(z)$ which would surely tell us that that the residue at $\infty$ is $-i$ (we aren't looking at the residue at $0$.) $\endgroup$
    – daruma
    Commented Aug 9, 2019 at 6:03
  • $\begingroup$ The residue at infinity is $-c_{-1}$ in the expansion around infinity. Or the residue of $-f(1/z)/z^2$ at zero. $\endgroup$
    – Maxim
    Commented Aug 9, 2019 at 6:24
  • $\begingroup$ Ah, right. I might have misunderstood that the residue of $f(1/z)$ at $0$ as the residue of $f(z)$ at $\infty$ (where it should be $-f(1/z)/z^2$.) $\endgroup$
    – daruma
    Commented Aug 9, 2019 at 6:49

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