I'm interested in finding polynomials $P$ for which there exists another polynomial $Q$ such that $\forall \theta$:
$$\| P(e^{i\theta}) \|^2 + \| Q(e^{i\theta})\|^2 = P(e^{i\theta})P^*(e^{i\theta}) + Q(e^{i\theta})Q^*(e^{i\theta}) = k \>\>\> \text{for some $k\in\mathbb{R_+}$}$$
My conjecture is that for all polynomials we can find such a pair but I'm not sure how to prove/disprove it. It might help to think of the polynomials as Fourier series given that they're polynomials of $e^{in}$.
Next, since the conjugate of $e^{in\theta}$ is $e^{-in\theta}$, after we multiply out the polynomials we end up with a number of constant terms corresponding to multiplying the exponentials with their own conjugate, and none constant terms. Since the sum must always equal to a constant and the Fourier basis is orthogonal, then the none constant terms from the multiplication of $P$ must cancel with those of $Q$ which gives us the following constraints: (supposed the highest degree present in either of the polynomials is $d$)
$$P_dP_0^* = -Q_dQ_0^*$$ $$P_dP_1^* + P_{d-1}P_0^*= -Q_dQ_1^* - Q_{d-1}Q_0^*$$ $$P_dP_2^* + P_{d-1}P_1^* + P_{d-2}P_0^*= -Q_dQ_2^* - Q_{d-1}Q_1^* - Q_{d-2}Q_0^*$$ $$...$$ $$P_dP_{d-1}^* + P_{d-1}P_{d-2}^* + ... +P_{1}P_0^*= -Q_dQ_{d-1}^* - Q_{d-1}Q_{d-2}^* - ... -Q_{1}Q_0^*$$
Where $P_i$ and $Q_i$ are the coefficients of the ith degree term in $P$ and $Q$ respectively.
The result is true with any $k \ge ||P||_{\infty}^2=\max |P(e^{i\theta})|^2$ and the proof is classical as $G(\theta)=|P(e^{i\theta})|^2$ is a (non negative) trigonometric polynomial of degree $d=\deg P$ that satisfies $G \le k$ by the hypothesis on $k$; but then $H=k-G$ is a non negative trigonometric polynomial of degree $d$ (here we can assume wlog $d>0$ as the result is trivial for $d=0$), so by a classical result (will include proof below) there is $Q$ a polynomial of degree $d$ st $H(\theta)=|Q(e^{i\theta})|^2$ and we are done!
Proof of the fact that if $H(\theta) \ge 0$ for all $\theta$ where $H$ is a degree $d$ trigonometric polynomial then $H(\theta)=|Q(e^{i\theta})|^2$ for some polynomial of degree $d$ (which can be taken to have no zeroes inside the open unit disc)
Note that $H(\theta)=e^{-in\theta}R(e^{i\theta})$ for some polynomial $R$ of degree $2d$ and since $H =\bar H$ we have $z^{2d}\overline {R(\frac{1}{\bar z})}=R(z)$ for all $|z|=1$. By the identity theorem this holds then for all $z \in \mathbb C$ since both sides are polynomials of degree $2d$, hence the roots of $R$ are either on the unit circle, or grouped in pairs $(w, \frac{1}{\bar w}), |w|>1$ say.
In the latter case, one has $(e^{i\theta}-w)(e^{i\theta}-\frac{1}{\bar w})=-\frac{e^{i\theta}}{\bar w}|e^{i\theta}-w|^2$, so it follows by grouping the roots $(w, \frac{1}{\bar w}), |w|>1$ together, we get $R(e^{i\theta})=c\Pi|e^{i\theta}-w|^2|\Pi(e^{i\theta}-e^{i\theta_j})$ or $H(\theta)=|Q_1(e^{i\theta})|^2H_1(\theta)$ where $Q_1$ is a polynomial of degree the number of pairs $(w, \frac{1}{\bar w}), |w|>1$ as above (with roots $|w|>1$) and $H_1$ is a nonnegative trigonometric polynomial with roots only on the unit circle, so we need to prove the result in the case that $R_1$ constructed as above for $H_1$ has all the roots on the unit circle.
But then by definition $H_1(z)=\frac{R_1(z)}{z^{d_1}}$ on the unit circle, hence extending $H_1$ to a rational function by the above equation, we get that $H_1$ has roots only on the unit circle and maps it into $[0, \infty)$; but then each root must have even multiplicity (otherwise $H_1$ maps a small neighborhood of it on the unit circle, into a small neighborhood of $0$ hence takes negative values too), so $R_1$ has all roots of even multiplicity, or $R_1=Q_2^2$ for some polynomial $Q_2$ and then since $H_1(\theta) \ge 0$ we get that $H_1(\theta)=|H_1(\theta)|=|Q_2^2(e^{i\theta})|/1=|Q_2(e^{i\theta}|^2$ and finally $H(\theta)=|(Q_1Q_2)(e^{i\theta})|^2$ so we are done!
Per comments here are two references for the proof above:
T Sheil-Small Complex Polynomials
QI Rahman and G Schmeisser Analytic Theory of Polynomials