$ z^3+z^2+2$ doesn't...

But $ z^3+z^2+2$ is not of the form $ az^{2n+1}+bz^{2n}+\bar bz+\bar a$. If you want $ a=b=1,n=1$ then $ \bar a=\bar b=1$ and your polynomial is $ z^3+z^2+z+1$ which does have a root (in fact it has all roots) on the unit circle.

Hi nayel, I think the problem was edited after rofler's post. Could you please check if this solution is correct?

Akashnil wrote:

I noticed that the last part isn't clear enough. Let me try to explain there a bit

I think it's simpler than that. If $ p(z)=0$ then $ z \neq 0$ since $ a \neq 0$ so we can write \[ p(1/\bar z) = a ( 1 / {\bar z}^{2n+1}) + b (1/{\bar z}^{2n}) + {\bar b}(1/{\bar z}) + {\bar a} = \frac{1}{ {\bar z}^{2n+1}} (a + b{\bar z} + {\bar b}{\bar z}^{2n} + {\bar a}{\bar z}^{2n+1}) = \frac{1}{ {\bar z}^{2n+1}} {\overline{p(z)} = 0}\] so the roots of $ p(z)$ can be paired up by matching each root $ z$ with $ \frac{1}{\bar z}$. Since the number of roots is odd, there must be a root paired with itself, meaning $ z = \frac{1}{\bar z}$ or $ |z| = 1$.

to randomgrapgh : how do u know that $ p$ doesn't have a repetetive root ?

Toxic_Frog wrote: to randomgrapgh : how do u know that $ p$ doesn't have a repetetive root ? I don't actually know that, and you are right: the argument I presented is incomplete in the case of repeated roots, since all I showed was that if $ z_0$ is a root, so is $ 1/\overline{z_0}$. This doesn't show if $ z_0$ is a root of multiplicity $ m$, so is $ 1/\overline{z_0}$. But this is actually the case. In fact, let $ f(z)$ be any polynomial of degree $ n$, not necessarily symmetric, and let $ \alpha_1, \ldots, \alpha_n$ be its roots. So we can write $ f(z) = a(z-\alpha_1)(z-\alpha_2)\cdots (z-\alpha_n)$, for some $ a$ which is the coefficient of $ z^n$. Consider the function $ \tilde{f}(z) = z^n \overline{f(1/\bar{z})}$ An easy calculation shows that $ \tilde{f}$ is actually a polynomial, and its roots are precisely $ \beta_i = 1/\overline{\alpha_i}$ - in other words, if $ \alpha_i$ occurs with multiplicity $ m$ in $ f$, then $ \beta_i$ occurs in multiplicity $ m$ in $ \tilde{f}$. In our case, $ p(z) = \tilde{p}(z)$, so they have the same roots with the same multiplicities. Therefore the pairing works as I described and one root must necessarily be paired with itself. Thanks for catching this error.

hello I wonder if this works Let $ z = e^{i \theta}$ as $ \theta$ runs from $ 0$ to $ 2\pi$ so we may uniquely define $ \sqrt {z} = e^{i \theta / 2}$. Then the condition that the polynomial have a root for some $ z$ of the form above is equivalent to $ a e^{i (n + 1/2) \theta} + b e^{i (n - 1/2) \theta} + \overline{b} e^{ - i (n - 1/2) \theta} + \overline{a} e^{ - i (n + 1/2) \theta}$, or that $ a e^{i (n + 1/2) \theta} + b e^{i (n - 1/2) \theta}$ is pure imaginary, or that its square $ (az + b)^2 z^{2n - 1}$ is nonpositive real. However, since this expression has at least $ 2n - 1$ roots within the unit circle (possibly more depending on the norm of the ratio of $ a$ and $ b$), as $ z$ traces the unit circle, the image curve must have winding number at least $ 2n - 1$ about the origin, and in particular must intersect the nonpositive real axis at least $ 2n - 1$ times. Hence the polynomial in question must have at least $ 2n - 1$ roots on the unit circle. I am concerned since this seems to be a much stronger result than desired.

What an interesting problem! Here is a (slightly) different way to work the same magic that is done in post #6. Suppose $z$ is a complex number such that $az^{2n+1}+bz^{2n}+\overline bz+\overline a=0$. Note that as $p(x)$ is equal to $0$, its complex conjugate $\overline{p(x)}$ is also equal to $0$. Hence we may write $\overline a \overline z^{2n+1}+\overline b\overline z^{2n}+b\overline z+a=0$. As $z\neq 0$, dividing by $\overline z^{2n+1}$ gives \[\overline{a}+\overline{b}\left(\dfrac1{\overline z}\right)+b\left(\dfrac1{\overline z^{2n}}\right)+a\left(\dfrac1{\overline z^{2n+1}}\right)=0\implies p\left(\dfrac1{\overline z}\right)=0.\] Therefore $p(z)=0$ implies $p(\tfrac1{\overline z})=0$, so each root can be paired with another unique root. But by the Fundamental Theorem of Algebra, the number of roots is odd. The only way this can happen is if there exists a complex number $z_0$ such that \[z_0=\dfrac1{\overline{z_0}}\implies z_0\overline{z_0}=|z_0|^2=1\implies |z_0|=1,\] as desired. $\blacksquare$