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Let $P(x)$ be a monic real valued polynomial with degree greater than $0$. If $|P(i)|<1$, prove that $P(x)$ has atleast a pair of complex roots. [$i= \sqrt{-1}$]


Let $$P(x)=x^n+a_1 x^{n-1}+a_2 x^{n-2}+\cdots+a_0$$ Suppose all the roots are real, then $$P(x)=(x-b_1)(x-b_2)\cdots (x-b_n)$$ where $b_j$ are real numbers. Also by Vieta's Relations, $$\sum_{j=1} ^n b_j = -a_1$$ I still haven't used the fact that $|P(i)|<1$. How to proceed from here?

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    $\begingroup$ @Digitallis your counter example has $|P(i)|=1$ and not $<1$ as OP requires $\endgroup$
    – Fotis
    Commented May 31, 2023 at 22:15

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For degree $1$ monic polynomial $P(x)$ with real coefficients, it's easy to see that we can never have $|P(i)|<1$.

So let $P(x)$ be a real polynomial of degree $n\geq 2$ such that $|P(i)|<1$ and let $b_1,\dots b_n$ be its roots.
Then we can write: $$P(x)=(x-b_1)\cdots(x-b_n)$$

We have: $$|P(i)|<1\Rightarrow |(i-b_1)\cdots(i-b_n)|<1\Rightarrow|i-b_1|\cdots|i-b_n|<1$$ This means that for some (plural) $b_k$ we get $|i-b_k|<1$ and for the rest $|i-b_k|>1$.
Obviously, we can't have $|i-b_k|>1$ for all $k$, since then their product would also be greater than $1$.

So there's at least one $j$, such that: $|i-b_j|<1$.

But then: $$|i-b_j|<1\Rightarrow \sqrt{1+b_j^2}<1\Rightarrow b_j^2+1<1\Rightarrow b_j^2<0$$

This is only possible if $b_j$ is complex, and since complex roots always come in conjugate pairs (for polynomials with real coefficients) we conclude that $P(x)$ has at least one pair of complex roots.

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    $\begingroup$ Slightly simpler: If $P(x)=(x-b_1)\cdots(x-b_n)$ with real $b_j$ then $|i - b_j| = \sqrt{1+b_j^2} \ge 1$ for all $j$ and therefore $|P(i)| \ge 1$. $\endgroup$
    – Martin R
    Commented Jun 5, 2023 at 7:00
  • $\begingroup$ @MartinR oh nice, and this covers the case of $\deg(P(x))=1$ as well, without having to separate it from the other. $\endgroup$
    – Fotis
    Commented Jun 5, 2023 at 17:02

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