Prove: $z^{12}+3z^8+101z^4+1$ has a root on the unit circle $\newcommand{\cis}{\operatorname{cis}}$>Prove that $$f(z)=z^{12}+3z^8+101z^4+1$$ has a root on the unit circle or $|z|\leq 1$
So started with looking at $$z^{12}+3z^8+101z^4+1=0$$
Therefore
$$z^{12}+3z^8+101z^4=-1$$
looking at $z=r\cis\theta$ we get
$$r^{12}\cis(12\cdot\theta)+3r^8\cis(8\cdot \theta)+101r^4\cis(4\cdot \theta)=-1$$
And I can see that if $x=r^4\cis(4\theta)$ we get
$$x^3+3x^2+101x+1=0$$
I also know that if $z$ is a solution so is  $\overline{z}$
How should I continue? 
Moreover: Can we say that $12$ degree polynomial as $12$ complex roots ($z$ and $\overline{z}$) but because we have $z^8$ and $z^4$ so there will be less than $12$ solutions?

You can apply Rouché's theorem. Let $f(z)=z^{12}+3z^8+101z^4+1$ and $g(z)=101z^4$. 
For $|z| \leq 1$ : 
$$|f(z)-g(z)|=|z^{12}+3z^8+1|\leq|z^{12}|+|3z^8|+|1|=5$$
$$|g(z)|=101$$
We can apply the theorem because $5<101$.
So $f$ has the same numbers of roots as $g$ in $\{z\in \mathbb{C};|z|\leq1\}$, so $f$ has $4$ roots in the domain considered.
