contour integration - $f(z)=sqrt{1-z^2}$ 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?