$\arg(z-R)\approx \pi$ is close to $0 \pmod{\pi}$. (No confusion should arise between the root $R$ and the curve $R=\{z{\;:~}{\mathop{\rm Re}\nolimits}f(z)=0\}$, which does not appear in this section.)

\[cubic-example\] Let $f(z)=z^3+iz^2+z-2$, a cubic whose associated basketball is shown on the left of Figure \[insert-example-figure\]. Every root of $f(z)$ has complex magnitude $<2$. If we choose a new root that is much larger in magnitude, say $R=8+8i$, then the basketball of $f(z)(z-R)$ is given combinatorially by “inserting an ear at $R$”, as shown.

The following theorem explains how ${M}(f(z)(z-R),\theta)$ is obtained from ${M}(f,\theta)$, and is the crucial result from which Theorem \[IBT\] will follow. We first introduce some useful notation. If $M$ is a matching on $[0,2n-1]$, define a matching $\widehat{M}$ on $[0,2n+1]$ by $$\widehat{M} = \big\{ \{0,2n+1\} \big\} \;\cup\; \big\{ \{i+1,j+1\} {\;:~}\{i,j\} \in M \big\}.$$ Also, for any positive real $R$, let $g_R(z) = (z-R)f(z)$.

\[thm:matchings\] Let $f(z)$ be a monic polynomial, and let $\theta \in (0,\pi)$ such that ${C_{\theta}}(f)$ is nonsingular. Then, for $R$ sufficiently large, we have $${M}(g_R,\theta) = \widehat{{M}(f,\theta)}.$$

Before we can prove Theorem \[thm:matchings\], we must develop several subsidiary results. The first of these is the following, which is implicit in the statement of Theorem \[thm:matchings\].

\[prop:smooth\] 