# Prove the polynomial has at least one complex root

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?

• @Digitallis your counter example has $|P(i)|=1$ and not $<1$ as OP requires May 31 at 22:15

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.

• 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 $|P(i)| \ge 1$. Jun 5 at 7:00
• @MartinR oh nice, and this covers the case of $\deg(P(x))=1$ as well, without having to separate it from the other. Jun 5 at 17:02