# $f(z)=\sqrt{1-z^2}$ pole at infinity

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?

• 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 $\int_{|z|=2} f(z)dz = -\int_{|s|=1/2} \frac{f(1/s)}{-s^2}ds = 2i \pi Res(\frac{f(1/s)}{s^2},0)$ Aug 8, 2019 at 22:23

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 $$\operatorname{Res}_{z = \infty} f(z) = 1/(2 i)$$.
• 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 $0$.) Aug 9, 2019 at 6:03
• The residue at infinity is $-c_{-1}$ in the expansion around infinity. Or the residue of $-f(1/z)/z^2$ at zero. Aug 9, 2019 at 6:24
• 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 $-f(1/z)/z^2$.) Aug 9, 2019 at 6:49