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We prove the claim. The change $z=\lambda w$, followed by division of $P$ by $\lambda^4$, multiplies the renormalized length by $\lambda$. It therefore suffices to treat

$$|a|\le1,\qquad |b|\le1,\qquad \max(|a|,|b|)=1. \tag{1}$$

The real constant $c$ is unrestricted. All constants below are independent of it.

The projection inequality

For $\alpha\in\mathbb R/\pi\mathbb Z$ and $s\in\mathbb R$ let

$$N(\alpha,s)=\#\{t\in\mathbb R: \operatorname{Re}P(e^{i\alpha}(t+is))=0\}.$$

Except for finitely many directions and a null set of tangent lines, this is a number between zero and four. If $A$ is any measurable set of such lines, the projection inequality is

$$J(P)\le-\frac12\int_A(4-N(\alpha,s))\,ds\,d\alpha. \tag{2}$$

Here $J(P)$ exists: centering gives eight ends with positions $re^{i(\pi/8+j\pi/4)}+O(r^{-1})$ and tangent derivatives differing from those of rays by $O(r^{-2})$. Each end therefore has a finite length excess over its radial coordinate. The compact portion of the real algebraic curve has finite length.

For clarity, (2) is the normalization of the projection argument in the proof of Fryntov and Nazarov's Lemma 1, pp. 5--6 of the preprint identified on the source card. Project the plane centrally onto the lower hemisphere of a sphere of radius $R$ tangent to it at zero. Spherical Crofton gives projected total length $4\pi R-D_R$, with

$$D_R=\frac12\int_0^\pi\int_{\mathbb R} (4-N(\alpha,s))(1+s^2/R^2)^{-3/2}\,ds\,d\alpha.$$

For all sufficiently large $r$, every circle of radius $\rho\ge r$ meets $H$ at least eight times, by the leading term $\operatorname{Re}z^4$. The radial coordinate on the hemisphere therefore bounds the length of the projected part outside $D_r$ from below by $4\pi R-8R\arctan(r/R)$. Its part inside has length at most $8R\arctan(r/R)-D_R$. Letting $R\to\infty$, restricting the nonnegative deficit integral to $A$, and then letting $r\to\infty$ gives (2). Equivalently the unoriented Euclidean line measure is $\tfrac12\,ds\,d\alpha$, which fixes the factor in (2).

A quadratic coefficient bounded away from zero

Suppose $|a|\ge1/100$. There is a direction

$$\alpha_0\in\{-\pi/12,\pi/12,\pi/2-\pi/12,\pi/2+\pi/12\}$$

for which $\cos4\alpha_0=1/2$ and $\operatorname{Re}(ae^{2i\alpha_0})\ge|a|/2$. Indeed, writing $a=x+iy$, the first two choices yield $\sqrt3x/2\pm y/2$, and the other two change their signs. The largest absolute value of the first two is at least $\sqrt{3x^2/4+y^2/4}\ge|a|/2$.

For $|\alpha-\alpha_0|\le1/100$, the elementary Lipschitz estimates for the exponential give, with $A_4=\cos4\alpha$ and $B=\operatorname{Re}(ae^{2i\alpha})$,

$$A_4\ge\frac12-\frac4{100}>\frac25, \qquad B\ge\frac{|a|}{2}-\frac{2|a|}{100}>\frac{2|a|}{5}. \tag{3}$$

In the real polynomial $q(t)=\operatorname{Re}P(e^{i\alpha}(t+is))$ the leading, cubic and quadratic coefficients are respectively

$$A_4,\qquad -4s\sin4\alpha,\qquad B-6s^2A_4.$$

After division by $A_4$ and removal of the cubic term, the quadratic coefficient is exactly

$$\frac{B}{A_4}-\frac{6s^2}{A_4^2}. \tag{4}$$

For $|s|\le\sqrt{|a|}/10$ it is positive, since $A_4B>4|a|/25>6|a|/100$ by (3). A monic real quartic with four real roots and zero cubic coefficient has quadratic coefficient $-\tfrac12\sum_j x_j^2\le0$. Thus $q$ has a nonreal conjugate pair, and $N(\alpha,s)\le2$ throughout this rectangle, apart from irrelevant null sets.

The rectangle has angular width $1/50$ and offset width $\sqrt{|a|}/5$. Substitution in (2) gives

$$J(P)\le-\frac{\sqrt{|a|}}{250}\le-\frac1{2500}. \tag{5}$$

Angles are taken modulo $\pi$; crossing an endpoint of a fundamental interval does not change the rectangle's line measure.

A substantial linear coefficient

It remains to treat $|a|<1/100$, so (1) gives $|b|=1$. Choose $\alpha_0\in\{0,\pi/2\}$ such that $|\operatorname{Re}(be^{i\alpha_0})|\ge1/\sqrt2>7/10$. For $|\alpha-\alpha_0|\le1/100$ and $|s|\le1/100$, write

$$q(t)=A_4t^4+d t^3+e t^2+f t+g.$$

Its coefficients satisfy

$$\begin{aligned} &\frac9{10}<A_4\le1,\qquad |d|\le\frac1{25}, \qquad |e|\le\frac1{100}+\frac6{10000}=\frac{53}{5000},\\ &|f|\ge\frac{69}{100}-\frac2{10000}-\frac4{10^6} >\frac23. \end{aligned} \tag{6}$$

Here $d=-4s\sin4\alpha$, $e=\operatorname{Re}(ae^{2i\alpha})-6s^2A_4$, and $f=\operatorname{Re}(be^{i\alpha}+2ias e^{2i\alpha}-4is^3e^{4i\alpha})$. The constant coefficient plays no role.

Divide by $A_4$ and remove the cubic term. The resulting quadratic and linear coefficients, $B_0$ and $C_0$, obey

$$\begin{aligned} |B_0|&\le\frac{|e|}{A_4}+\frac{3d^2}{8A_4^2}<\frac1{50},\\ |C_0|&\ge\frac{|f|}{A_4} -\frac{|de|}{2A_4^2}-\frac{|d|^3}{8A_4^3}>\frac12. \end{aligned} \tag{7}$$

These inequalities follow directly by inserting (6); the two error terms in the second line sum to less than $1/1000$. If the depressed quartic had four real roots $x_j$, Newton's identities would give

$$\sum_jx_j=0,\qquad \sum_jx_j^2=-2B_0\le\frac1{25}, \qquad \left|3C_0\right|=\left|\sum_jx_j^3\right| \le\left(\sum_jx_j^2\right)^{3/2}\le\frac1{125}.$$

This contradicts (7). Hence again $N(\alpha,s)\le2$. The angular and offset widths are both $1/50$, so (2) gives $J(P)\le-1/2500$. Together with (5), this proves the normalized estimate. Restoring the scale $\lambda$ proves the claim.