= Solution
The <dominant balance for algebraic roots> separates one small <root of a polynomial> from three large ones. For the small <root of a polynomial>, write $z=\varepsilon+a\varepsilon^3+O(\varepsilon^5)$ in $z=\varepsilon(1+z^2)^2$. Comparing coefficients gives $a=2$, so
$$
\boxed{z_{\mathrm{small}}=\varepsilon+2\varepsilon^3+O(\varepsilon^5).}
$$
This is a real <analytic function> of real $\varepsilon$ near zero, since the derivative with respect to $z$ of $z-\varepsilon(1+z^2)^2$ is one at $(0,0)$.
For the large <roots of a polynomial>, use the real cube root $s=\operatorname{sgn}(\varepsilon)|\varepsilon|^{1/3}$ and set $z=w/s$. The transformed equation is
$$
w=(w^2+s^2)^2.
$$
Its three nonzero leading <roots of a polynomial> are $w=\omega$, with $\omega^3=1$. Writing $w=\omega+as^2+O(s^4)$ gives $a=4a+2\omega^2$, hence $a=-2\omega^2/3$. Therefore
$$
\boxed{z_\omega=\frac{\omega}{s}-\frac23\omega^2s+O(s^3),\qquad \omega=1,e^{2\pi i/3},e^{-2\pi i/3}.}
$$
The choice $\omega=1$ is the second real <root of a polynomial>; the other two form a <complex conjugate> pair for either sign of $\varepsilon$. The nonzero transformed <roots of a polynomial> are simple, so their <Taylor series> in $s^2$ exist locally and exhaust the three large <roots of a polynomial>. Together with the small one these give all four <roots of a polynomial> for sufficiently small nonzero $\varepsilon$. At $\varepsilon=0$ itself the rational equation has only the finite root zero; the other three escape to infinity. Multiplying by $(1+z^2)^2$ has not introduced roots at $z=\pm i$, since those points do not solve the multiplied equation.
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