two complex polynomials with equal modulus on a parabola Are there any two complex polynomials $P(z)$ and $Q(z)$ with the same degree, such that their modulus is equal on the  parabola $y=x^2$ but they are not a  constant  multiple of each other? in other words, can we have $|p(x+ix^2)|=|q(x+ix^2)|$ for two complex polynomials $p$ and $q$ for all real $x$, where $p\not=q.$
The above quesion is equivalent to the following:
is there any rational complex fuction sending the parabola $y=x^2$ to the unit circle?
 A: There is no nonconstant rational function taking the parabola to the unit circle. Proof: Consider any rational function of one complex variable. It extends to a smooth map $\phi: \mathbb{CP}^1 \to \mathbb{CP}^1$. Let $\gamma$ be the closure of the parabola; it is a closed curve in $\mathbb{CP}^1$. We claim that $\phi(\gamma)$ is not smooth and, in particular, is not the unit circle.
Let's first see what $\gamma$ looks like near $\infty$. Geometrically, the two branches of the parabola come into $\infty$ in the same direction, so $\gamma$ has a cusp at $\infty$. If you don't trust your geometric intuition, here is a direct computation. Let $u+iv$ be a coordinate near $\infty$, so $x+iy = (u+iv)^{-1} = u/(u^2+v^2) - i v/(u^2+v^2)$. So the equation of $\gamma$ is 
$$\frac{-v}{u^2+v^2} = \left( \frac{u}{u^2+v^2} \right)^2$$
or, clearing denominators,
$$(u^2+v^2) v + u^2 =0$$
This cubic has a cusp at $(0,0)$, as shown in the figure below.
       (source)
Now, let $e$ be the order of branching of $\phi$ at $\infty$. So $\phi$ multiplies angles by the integer $e$. We see that the curve $\phi(\gamma)$ will also come into $\phi(\infty)$ in the same direction from both sides. In particular, $\phi(\gamma)$ is not smooth at $\phi(\infty)$, and cannot be the unit circle.
A: If P and Q have the same degree and leading coefficient, then P/Q converges to 1 at infinity. Consequently, the function
$$g(w)=\log(P(1/w)/Q(1/w))$$
has a single valued branch near the origin. The real part of g would have to be zero on the curve $(u^2+v^2)v=-u^2$, where $w=u+iv$. This curve has a cusp at the origin, so it cannot be part of the zero set of a harmonic function.
A: Yes, take $P$ and $Q=e^{it}P$.
