# Finding a lower bound on the roots

Question Statement-

If $$z$$ is a complex number and $$x_0, x_1,..., x_{2n}$$ are real then prove that all the roots of the equation of $$(sinx_0)z^{2n}+(cosx_1)z^{2n-1}+(sinx_2)z^{2n-2}+...+(cosx_{2n-1})z+sinx_{2n}=b$$ lie in the region $$|z|>1-\frac{1}{|b|}$$ My working- I tried using the conventional methods of vieta's relations and taking conjugates and somehow arrive on some conclusions. Then I tried proving that maybe all the roots have to be real or all the root's complex conjugate must also be a root, as only possible complex coefficient is $$b$$. I have notice that the inequality essentially means we have to prove $$|z|>1$$ and also all the sin and cos functions essentially imply that all coefficients are less than or equal to 1 in magnitude. But I haven't gotten much out of it. Would love some hints!

Hint: assume for the same of contradiction that $$z$$ is a root of the polynomial equation and $$|z| \le 1-1/|b|$$. Find an upper bound for the left-hand side using the triangle inequality and the sum of a geometric series.

• Thanks! I tried using proof by contradiction by same claim but couldn't see how, thanks again for the help.
– bm27
Feb 21 at 7:38

The $$\cos$$ and $$\sin$$ here are not relevant, we can simplify to: $$\sum_{n=0}^N a_n \ z^n = b, \ \mbox{with all} \ |a_n| \leq 1$$ Then use that for $$|z|\geq 1$$ the claim it is trivially true and that for $$|z|<1$$ we can write: $$|b| = \left|\sum_{i=0}^{2n} a_i \ z^i\right| \leq \left|\sum_{i=0}^{2n} |a_i| \cdot |z|^i\right| \leq \left|\sum_{i=0}^{2n} 1 \cdot |z|^i\right| < \left|\sum_{i=0}^\infty |z|^i\right| \leq \frac{1}{1-|z|},$$ which gives the required result since there is one $$<$$ sign in the sequence.

• This is basically what Greg hinted me to do, anyways thanks for the answer
– bm27
Feb 21 at 9:27
• Yes Greg's answer showed up too late on my side... Feb 21 at 9:36
• yeah I think we were composing our answers simultaneously :) Feb 21 at 17:17