Proof

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Use the conventions of L186. Translation centers a monic quartic. The rotation $p(z)\mapsto e^{-4i\theta}p(e^{i\theta}z)$ preserves monicity and length, as does conjugation of all coefficients.

A compact parameter set

Write $p(z)=z^4+az^2+bz+c$ and let $u,v,w$ be its critical points, counted with multiplicity. Then

$$u+v+w=0,\qquad p'(z)=4(z-u)(z-v)(z-w),\qquad p(u)-p(v)=w(u-v)^3. \tag{1}$$

For the last identity integrate $p'$ from $v$ to $u$, writing $z=(u+v)/2+t$. The odd part vanishes, and the integral is $-(u+v)(u-v)^3$. The identity also holds when critical points coincide.

Let $M=\max(|u|,|v|,|w|)$ and label $|u|=M$. If $|v-w|\ge M/2$, then $|p(v)-p(w)|\ge M^4/8$. Otherwise $|v|,|w|>M/4$ and $|u-v|=|3u/2-(v-w)/2|>5M/4$, so $|p(u)-p(v)|>125M^4/256$. Consequently

$$\max_{i,j}|p(u_i)-p(u_j)|\ge M^4/8. \tag{2}$$

If all critical values have modulus one, (2) gives $M\le2$. Comparing coefficients in $p'$ now gives

$$|a|\le6M^2\le24,\qquad |b|\le4M^3\le32, \qquad |c|\le1+M^4+|a|M^2+|b|M\le177. \tag{3}$$

The centered coefficient set with all critical values on the circle is closed: along coefficient convergence the multiset of roots of $p'$ converges, and their polynomial values converge. It is bounded by (3), hence compact. More generally, critical values in $|z|\le K$ give $M^4\le16K$ and a coefficient bound depending only on $K$. This also supplies compactness when critical values are moved along fixed bounded arcs. No classification theorem for the critical-value map is needed for these bounds.

The generic two-real-parameter chart

Assume first that the critical points and the critical values are each pairwise distinct. Choose a critical point of largest modulus, and rotate so that it equals $s>0$. Write the other two as $st,-s(1+t)$, where

$$|t|\le1,\qquad |1+t|\le1.$$

Define

$$F_t(z)=z^4-2(1+t+t^2)z^2+4t(1+t)z.$$

Then $p(sz)=s^4F_t(z)+c$. Put $f_1=F_t(1)$, $f_2=F_t(t)$ and $f_3=F_t(-1-t)$. The three distinct critical values on a circle are not collinear. Thus the $f_i$ have a unique circumcenter $C(t)$ and positive circumradius $R(t)$. They can be obtained from the two real linear equations

$$2\operatorname{Re}\bigl((C-f_1)\overline{(f_i-f_1)}\bigr) =|f_i-f_1|^2,\qquad i=2,3.$$

The unit-circle condition forces $s^4R(t)=1$ and $c=-s^4C(t)$. Conversely these choices give unit critical values. Every member of the generic class is therefore, up to the stated isometries, of the form

$$p_t(z)=s^4\bigl(F_t(z/s)-C(t)\bigr),\qquad s=R(t)^{-1/4}. \tag{4}$$

The parameter domain is the part of the closed lens $|t|,|1+t|\le1$ where the three $f_i$ are noncollinear and the three critical points are distinct. Relabeling critical points can give several parameters for the same polynomial; injectivity is unnecessary for coverage. The chart itself is singular at excluded parameters. Compactness of the coefficient set does not give bounded derivatives of (4) there.

The missing strata are explicit

There are three additional cases, with complex conjugation allowed to identify the two signs below, since it preserves length.

A double critical point. The roots of $p'$ are $r,r,-2r$ with $r\ne0$. Rotate so that $r>0$ and put $h=r^4$. The polynomial is

$$p(z)=z^4-6\sqrt h\,z^2+8h^{3/4}z+ \frac{21h}{2}\ \pm i\sqrt{1-\frac{729h^2}{4}}, \qquad 0<h\le\frac2{27}. \tag{5}$$

Indeed its critical values are $c+3h$ and $c-24h$, whose difference is $27h$. Requiring both moduli to equal one gives (5). This family has a double critical point and a distinct simple one throughout $h>0$.

Distinct critical points with two equal critical values. By (1), if $u\ne v$ but $p(u)=p(v)$, the third critical point is zero. Thus the roots are $r,-r,0$. Rotate so that $r>0$ and again put $h=r^4$. The family is

$$p(z)=z^4-2\sqrt h\,z^2+\frac h2 \ \pm i\sqrt{1-\frac{h^2}{4}},\qquad 0<h\le2. \tag{6}$$

Its critical values are $c-h,c-h,c$, so the unit-circle condition gives (6). These are collisions of values at distinct points, and are not covered by considering repeated roots of $p'$ alone.

A triple critical point. Centering forces it to be zero, so $p(z)=z^4+c$ with $|c|=1$. Rotation makes this $z^4-1$.

These cases exhaust the complement of (4). Three distinct collinear critical values cannot lie on the unit circle. If all critical values equal $v_0$, each distinct critical point of multiplicity $m$ is a root of $p-v_0$ of multiplicity $m+1$. Their total multiplicity is three plus the number of distinct critical points, at most four. Thus there is just one critical point, the last case above.

Unicritical length comparison

Away from finitely many critical values, inverse branches give

$$\ell(p)=\int_0^{2\pi}\sum_{p(z)=e^{i\theta}}\frac{d\theta}{|p'(z)|}.$$

For $p(z)=z^4-c$ the sum is $|c+e^{i\theta}|^{-3/4}$. At $c=1$ substitution $u=\theta/2$ gives $\ell(z^4-1)=2^{1/4}B(1/2,1/8)$, with the usual beta integral.

For the single-critical-value class the required comparison is elementary. Put $\Phi(c)=\ell(z^4-c)$ for real $c\ge0$. For $c<1$, expanding the two factors of $|1+ce^{i\theta}|^{-3/4}$ and integrating gives

$$\Phi(c)=2\pi\sum_{j\ge0}\binom{-3/8}{j}^{\!2}c^{2j}.$$

It is strictly increasing on $[0,1)$. Dominated convergence at $c=1$ follows from $|c+e^{i\theta}|^2=(1-c)^2+c|1+e^{i\theta}|^2$ for $c\ge1/2$. For $c>1$, $\Phi(c)=c^{-3/4}\Phi(1/c)<\Phi(1)$. Rotating the constant term therefore shows that $\ell(z^4+c)\le\ell(z^4-1)$, with equality exactly when $|c|=1$.