polar coordinates $z=re^{i\theta}$, the curves $R$ and $I$ are given by equations $$\begin{aligned}
  {\mathop{\rm Re}\nolimits}\,f(z) &= r^n \cos n\theta + \text{(lower terms)} = 0,\\
  {\mathop{\rm Im}\nolimits}\,f(z) &= r^n \sin n\theta + \text{(lower terms)} = 0.
  \end{aligned}$$ Let $S_r$ denote the circle $\{z{\;:~}|z|=r\}$, and $D_r$ the disk $\{z{\;:~}|z|\le r\}$. By taking $r$ sufficiently large, we can ensure that

-   ${\mathop{\rm Re}\nolimits}\,f(z)$ has $2n$ zeros on $S_r$, arbitrarily close (in angle) to the zeros of $\cos n\theta$; and

-   ${\mathop{\rm Im}\nolimits}\,f(z)$ has $2n$ zeros on $S_r$, arbitrarily close to the zeros of $\sin n\theta$.

That is, $R \cap S_r$ and $I \cap S_r$ each contain exactly $2n$ points, and these $4n$ points alternate around the circle. (See Figure \[basketball-example-figure\] for an example.) Suppose that $R$ and $I$ are both smooth curves. Then the intersection $R\cap D_r$ must contain $n$ disjoint arcs $R_1,\ldots,R_n$, each of which joins two distinct points of $R
\cap S_r$. Since these arcs do not cross, each arc $R_i$ must have an even number of points of $R\cap S_r$ on either side of it, by the Jordan Curve Theorem.

Likewise, $I\cap D_r$ contains $n$ disjoint arcs $I_1,\ldots,I_n$, each of which joins two distinct points of $I \cap S_r$. Since 