Roots of a polynomial have unit modulus

Question

Is there a necessary and sufficient condition for a polynomial to have all roots of unit modulus?

Today, I came across the following problem, Prove that all the roots of the polynomial $11x^{10}+10ix^9+10ix-11=0$ satisfy $|x|=1$. I proceeded by putting $x=iy$ to get rid of the imaginary coefficients, and tried to use the fact that $f(x)=x^{10}f(\frac{1}{x})$, but couldn't do much after that. The solution proceeds as follows, $$|z|=\frac{|11-10iz|}{|11z^9+10|}$$ and then make cases $|z|<1$ or $|z|>1$ and arrive at a contradiction in both cases.

Answer 1

Working on what you did: After the substitution $x=iy$ , we get $11y^10+10y^9+10y+11=0$. Now, as $y=0$ is not a root of this equation, dividing both sides by $y^5$ gives $11y^5+11/y^5+10y^4+10/y^4=0$. Now put $t=y+1/y$ to get $11t^5+10t^4-55t^3-40t^2+55t+20=0$. You can plot the equation in $t$ to see that the equation has 5 real roots. So, every value of $t$ which satisfies the equation is real. As $t=y+1/y$, $y+1/y$ is also real. Taking $y=re^($i$\theta)$ you can see that $r$ must be 1 for$y+1/y$ to be real. As $r$ is the magnitude of $y$ ,the roots have unit modulus.