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roots of complex polynomial Let is $ p(z) = a_0 + ... + a_n z^n$ such that $a_n$ is not equal to zero there is $ j \in \{0, ... ,n \}$ with $ |a_j| > ( \sum_{k=0}^n |a_k|) - |a_j| $. Show that the polynomial $g(z) = p(z) - a_jz^j$ has not roots in $S^1 = \{ z; |z| = 1\}$.

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  • $\begingroup$ Sorry in problem about , we have that $ |a_j| > ( \sum_{k=0}^n |a_k|) - |a_k| $ $\endgroup$
    – user251833
    Jul 5, 2015 at 20:53
  • $\begingroup$ Did you perchance mean to write $\vert a_j \vert \le (\sum_{k = 0}^n \vert a_k \vert) - \vert a_j \vert$? $\endgroup$ Jul 5, 2015 at 20:55
  • $\begingroup$ No, sorry. In problem we have that $ |a_j| > ( \sum_{k=0}^n |a_k|) - |a_k| $. Sorry, I'm new here. $\endgroup$
    – user251833
    Jul 5, 2015 at 21:00
  • $\begingroup$ My real question concerns the presence of $\vert a_k \vert$ subtracted from the sum on the right; are you sure you don't want $\vert a_j \vert$ there? $\endgroup$ Jul 5, 2015 at 21:09
  • $\begingroup$ Now the problem is correct. $\endgroup$
    – user251833
    Jul 5, 2015 at 21:44

1 Answer 1

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Counterexample:

Let $p(z) = z^2 + 3z -1$. Then $|a_0| = 1, |a_1| = 3, |a_2| = 1$.

Clearly $3 = |a_1| > |a_0| + |a_2| = 2$, so $g(z) = z^2 - 1$.

But the roots of $g(z)$ are $\pm 1\in S^1$.

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