Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 76 / 1 / a

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 $z=\epsilon +a{\epsilon }^3+O({\epsilon }^5)$ in $z=\epsilon (1+z^2{)}^2$. Comparing coefficients gives $a=2$, soThis is a real analytic function of real $\epsilon$ near zero, since the derivative with respect to $z$ of $z-\epsilon (1+z^2{)}^2$ is one at $(0,0)$.

For the large roots of a polynomial, use the real cube root $s=\mathrm{sgn}(\epsilon )|\epsilon {|}^{1/3}$ and set $z=w/s$. The transformed equation isIts three nonzero leading roots of a polynomial are $w=\omega$, with ${\omega }^3=1$. Writing $w=\omega +a{s}^2+O(s^4)$ gives $a=4a+2{\omega }^2$, hence $a=-2{\omega }^2/3$. ThereforeThe choice $\omega =1$ is the second real root of a polynomial; the other two form a complex conjugate pair for either sign of $\epsilon$. 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 $\epsilon$. At $\epsilon =0$ 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.