 Solution

Definitions. In a punctured neighborhood of $z_0$, let the Laurent series be $f(z)={\sum }_{j=-\infty }^{\infty }{c}_j(z-z_0{)}^j$. The point is a pole of order $m\ge 1$ if $c_{-m}\ne 0$ and every coefficient with index less than $-m$ vanishes. Its residue is $\boxed{{\mathrm{Res}}_{z_0}f=c_{-1}}$. A simple pole has $m=1$.

Poles for $0<a<1$. Denote the bracket in the denominator by $D(z)$. Factoring givesPut $q=\sqrt{(1-a)/(1+a)}$, so $0<q<1$. The finite poles are $0,\pm q,\pm q^{-1}$, all simple; the numerator is nonzero at each. At zero,At a nonzero simple root $r$ of $D$, the residue formula givesFor $r^2=q^2$, $D^{\prime }(r)=8ar$; for $r^2=q^{-2}$, $D^{\prime }(r)=-8ar$. Substitution yieldsThe two zeros of a function at $z=\pm i$ do not cause additional poles: there $D=-4$ and the numerator vanishes.

The case $a=1$. Simplify before using any formulas that divide by $1-a^2$:Thus its only finite pole isAs $a\uparrow 1$, the two inner poles merge with zero while the outer two escape to infinity. If one also considers the Riemann sphere, $f$ at $a=1$ has a simple pole at infinity because $f(z)\sim z/4$; the requested unit-circle calculation only uses finite poles.

Unit-circle integral. Set $z=e^{i\theta }$. ThenAt the two angles where the tangent is undefined, the real integrand has continuous limiting value zero. Thus the substitution is valid for the improper integral, or after that continuous extension. No pole lies on $|z|=1$ for any allowed $a$.

For $0<a<1$, the sum of the enclosed residues isThe residue theorem therefore givesAt $a=1$ the enclosed residue is $1/2$, giving $I=\pi$, exactly the same formula. The direct simplification $1/(1+{\tan }^2\theta )={\cos }^2\theta$ provides an independent check of that endpoint.