If $~\vert a_r\vert<2,~~r=1,2,..n~$ then p.t. there is no $~z~$ inside the circle $~|z|=\frac{1}{3}~$ such that $~a_nz^n+a_{n-1}z^{n-1}..a_1z+1=0~$

Question

If $~~\vert a_r \vert < 2 ,~~~~r=1,2,3...n~~$ then prove that there is no $~z~$ inside the circle $~\vert z \vert = \frac{1}{3}~$ such that $$a_nz^n + a_{n-1}z ^{n-1}... a_1z +1 = 0$$

This is a question from complex numbers.

Answer 1

Try the triangle inequality. -

$$a_nz^n + a_{n-1}z ^{n-1}... a_1z = -1$$ Now take modulus.. $$|a_nz^n + a_{n-1}z ^{n-1}... a_1z |= 1$$ Apply inequality $$|a_nz^n + a_{n-1}z ^{n-1}... a_1z |<|a_nz^n| + |a_{n-1}z ^{n-1}|... |a_1z |$$ And as $|a_r| < 2$ $$ 1<|2 z^n| + |2z ^{n-1}|... |2z |$$ This is a GP now $$1 <\frac{2|z|(1-|z^n|}{1-|z|}$$ Rearrange $$\frac{1+2|z| ^n }{3} < |z|$$ And as $|z| > 0$ , $$\frac{1 }{3} < |z|$$

Initially I was unable to solve, but when the method popped up in my head, I thought it would be a good opportunity to self answer, so that anyone else could also benefit from this in the future.

Answer 2

This actually does not have much to do with complex numbers. Note that $$|\sum^{n}_{i=1}a_{i}z^{i}|\leq \sum^{n}_{i=1}|a_{i}z^{i}|<2\sum^{n}_{i=1}3^{-i}=1-3^{-n}<1.$$ Hence $$a_{n}z^{n}+a_{n-1}z^{n-1}+...+a_{1}z+1\neq0.$$