= Solution
\b[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 $m\geq1$ if $c_{-m}\ne0$ and every coefficient with index less than $-m$ vanishes. Its <residue> is $\boxed{\operatorname{Res}_{z_0}f=c_{-1}}$. A <simple pole> has $m=1$.
\b[Poles for $0<a<1$.] Denote the bracket in the denominator by $D(z)$. Factoring gives
$$
D(z)=[(a-1)z^2+(a+1)][(a+1)z^2+(a-1)].
$$
Put $q=\sqrt{(1-a)/(1+a)}$, so $0<q<1$. The finite <poles> are $0,\pm q,\pm q^{-1}$, all simple; the numerator is nonzero at each. At zero,
$$
\operatorname{Res}_0f=\frac1{a^2-1}=-\frac1{1-a^2}.
$$
At a nonzero simple root $r$ of $D$, the <residue> formula gives
$$
\operatorname{Res}_rf=\frac{(r^2+1)^2}{rD'(r)},\qquad D'(r)=4r[a^2(r^2+1)-(r^2-1)].
$$
For $r^2=q^2$, $D'(r)=8ar$; for $r^2=q^{-2}$, $D'(r)=-8ar$. Substitution yields
$$
\boxed{\begin{array}{c|c|c}\text{point}&\text{pole order}&\text{residue}\\\hline
0&1&-1/(1-a^2)\\
+q,-q&1&1/[2a(1-a^2)]\\
+q^{-1},-q^{-1}&1&-1/[2a(1-a^2)]
\end{array}\quad(0<a<1).}
$$
The two <zeros of a function> at $z=\pm i$ do not cause additional <poles>: there $D=-4$ and the numerator vanishes.
\b[The case $a=1$.] Simplify before using any formulas that divide by $1-a^2$:
$$
D(z)=(z^2+1)^2-(z^2-1)^2=4z^2,\qquad f(z)=\frac14\left(z+\frac2z+\frac1{z^3}\right).
$$
Thus its only finite <pole> is
$$
\boxed{z=0:\quad\text{order }3,\qquad\operatorname{Res}_0f=\tfrac12\quad(a=1).}
$$
As $a\uparrow1$, 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>.
\b[Unit-circle integral.] Set $z=e^{i\theta}$. Then
$$
\tan\theta=\frac{z^2-1}{i(z^2+1)},\qquad d\theta=\frac{dz}{iz},\qquad\frac{d\theta}{a^2+\tan^2\theta}=\frac1i f(z)\,dz.
$$
At 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> is
$$
-\frac1{1-a^2}+\frac1{a(1-a^2)}=\frac1{a(1+a)}.
$$
The <residue theorem> therefore gives
$$
\boxed{I=\frac1i\oint_{|z|=1}f(z)\,dz=\frac{2\pi}{a(1+a)},\qquad0<a\leq1.}
$$
At $a=1$ 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.
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