Proving that roots of a equation lies outside the circle $|z|=1/2$ 
Show that all the roots of the equation $$z^n \cos(\theta_0) + z^{n-1} \cos(\theta_1)+ \dots+\cos(\theta_n)=2$$ where $\theta_0,\theta_1,\dots,\theta_n \in \mathbb{R}$, lie outside the circle $|z|=1/2$.

I tried with the case when all theta are 0 then equation simplifies to $1+z+z^2+\dots+z^n=2$ . We know that if $1+z+z^2+\dots+z^n=0$ then $z_i$ are the complex $(n+1)$th roots of unity. So in this case origin is shifted such that here $z_i$ are roots of equation $$\left(z-\frac{2}{n+1}\right)^{n+1}=1.$$ I couldn't proceed after. this Any ideas?
 A: Suppose to the contrary that $z$ is a root of the above equation and that $|z|\le 1/2$. Since $|\cos(\theta_k)| \le 1$ we get
$2 \le \frac{1}{2^n}+\frac{1}{2^{n-1}}+...+\frac{1}{2}+1=2-\frac{1}{2^{n}}<2$,
a contradiction.
A: $$2=S_n=|z^n \cos(\theta_0) + z^{n-1} \cos(\theta_1)+ \dots+\cos(\theta_n)|$$
Using this,
$$S_n\le\sum_{r=0}^n|z^r||\cos\theta_r|\le\sum_{r=0}^n|z^r|$$
Now if $|z|\le\dfrac12,$  
$$\sum_{r=0}^n|z^r|\le\sum_{r=0}^n\left(\dfrac12\right)^r=2\left(1-\left(\dfrac12\right)^{n+1}\right)<2$$
A: Suppose that $|z|\leq \frac{1}{2}$. Then $|z|^i\leq \frac{1}{2^i}$ for all $i\in\mathbb{N}$. 
If
$$z^n \cos(\theta_0) + z^{n-1} \cos(\theta_1)+ \dots+\cos(\theta_n) =2,$$
then certainly
$$|z^n \cos(\theta_0) + z^{n-1} \cos(\theta_1)+ \dots+\cos(\theta_n)|=2.$$
By the triangle inequality,
$$\begin{align}|z^n \cos(\theta_0) + z^{n-1} \cos(\theta_1)+ \dots+\cos(\theta_n)|&\leq |z^n\cos(\theta_0)|+\ldots+|\cos(\theta_n)|\\
&\leq |z^n|+|z^{n-1}|+\ldots+1\\
&\leq \frac{1}{2^n}+\ldots+1\\
&=2-\frac{1}{2^n}\\
&<2
\end{align}$$
Therefore, there are no solutions $z$ to the equation such that $|z|\leq \frac{1}{2}$.
