# If $z_0$ is a root of the equation $z^n\cos\theta_0+z^{n-1}\cos\theta_1+\cdots+\cos\theta_n=2$

If $$z_0$$ is a root of the equation $$z^n\cos\theta_0+z^{n-1}\cos\theta_1+\cdots+\cos\theta_n=2$$, then

1. $$|z_0|<1/2$$

2. $$|z_0|>1/2$$

3. $$|z_0|=1/2$$

Using triangle inequality,$$|z^n\cos\theta_0|+|z^{n-1}\cos\theta_1|+\cdots+|\cos\theta_n|\ge2$$

$$|\cos\theta|\le1$$

Inequality won't change if every cosine takes its maximum value.

$$|z|^n+|z|^{n-1}+\cdot+1>2$$ I tried using GP formula but I got the correct answer only when I assumed $$|z|<1$$ and took an infinite GP. But I don't know about $$|z|$$.

You know that $$2< |z|^n+|z|^{n-1}+\cdot+1 =\frac{1-|z|^{n+1}}{1-|z|}$$
Now, if $|z| <1$ we have $$1-|z|^{n+1} > 2-2|z| \Rightarrow 2|z|>1+|z|^{n+1}>1$$
If $|z|\geq 1$ we have $$|z|>1 \geq \frac{1}{2}$$
P.S. Alternately, you can observe that if $|z| \leq \frac{1}{2}$ then $$1+|z|+|z|^2+..+|z|^n \leq 1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n} <2$$