# 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?

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Erm. Isn't it "moduli"? – Koundinya Vajjha Jun 6 '11 at 16:28
maybe you should include that they have leading coefficient $1$. – Olivier Bégassat Jun 6 '11 at 16:43

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.

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.

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It seems true, but I could not follow. The one below seems to use the same idea (Michael's solution) – mosen Jun 8 '11 at 9:58
Would it help if I linked you to the wikipedia entry for Riemmann sphere? en.wikipedia.org/wiki/Riemann_sphere – David Speyer Jun 8 '11 at 14:54
Can you explain what order of branching is, or intrudude a reference for that? Do you suggest that a smooth function should preserve the angles? – mosen Jun 24 '11 at 15:26
Almost. Let $f$ be a nonconstant analytic function defined near a point $z_0$. Change variables so that $z_0 = f(z_0) = 0$. So $f = c_e z^e + c_{e+1} z^{e+1} + \cdots$, for some positive integer $e$ and some nonzero $c_e$. If $e=1$, then $f$ preserves angles near $z_0$. In general, $f$ multiplies angles by a factor of $e$ at $z_0$. This $e$ is the order I am referring to. – David Speyer Jun 24 '11 at 15:45

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.

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Why can't a curve with a cusp at origin be the zero set of a harmonic function? – mosen Jun 16 '11 at 11:28

Yes, take $P$ and $Q=e^{it}P$.

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