Suppose $p(z)=z^2+az+b$ for some $a,b$ complex numbers, and $|p(z)|=1$ whenever $|z| = 1$.
Prove that $a=b=0$.
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(For a general result see Characterizing nonconstant entire functions with modulus 1 on the unit circle which proves that the only entire functions that satisfy $|f(z)|=1$ on the unit circle $|z|=1$ are polynomials of the form $f(z)=a z^n$ with $|a|=1$. The following is an elementary proof, not using complex analysis, for $2^{nd}$ degree polynomials such as the one in the posted question.)
$$|z|=1 \iff |z|^2 = z \bar z = 1 \iff \bar z = \frac{1}{z}$$
$$|p(z)|=1 \iff |p(z)|^2=1=p(z) \overline{p(z)}=p(z) p(\bar z)=p(z) p(\frac{1}{z})$$
Substituting $p(\frac{1}{z}) = \frac{1}{z^2}(1 + a z + b z^2)$ gives:
$$ (z^2 + az + b)(1 + a z + b z^2) = z^2$$
$$q(z) = b z^4 + a(b+1)z^3 + (a^2+b^2)z^2+a(b+1)z + b = 0$$
But the latter shows that the polynomial $q(x)$ has all points on the unit circle $|z|$ as roots, thus it has infinitely many roots, which can only happen if $q \equiv 0$. Equating coefficients to $0$ gives $a=b=0$.
