Say $P$ is a polynomial of degree $n$ with $n$ real roots. I have to prove that the polynomial $Q(z)=P(z+i)+P(z-i)$ has $n$ real roots.

First it happens that for a real number $t$ and because $P$ is real polynomial we have $$\overline{Q}(t)= \overline{Q(t)} =\overline{P(t+i)}+\overline{P(t-i)} \\ = P(\overline{t+i})+P(\overline{t-i}) \\ = P(t-i)+P(t+i) = Q(t) $$ The polynomial $Q-\bar{Q}$ has infinite roots (namely all the real numbers) and thus $Q$ is a real polynomial

Truth is I don't know how to proceed further. This exercice was given at the beginning of a complex analysis course so I don't know if I must search for an analytical argument or a simpler one. If you can just give me a hint (I want to solve the problem by myself) that would be really helpful

  • $\begingroup$ Why do you think that $\overline{P(t+i)}=P\left(\overline{t+i}\right)$? $\endgroup$ Nov 19, 2020 at 23:00
  • $\begingroup$ $P$ is a real polynomial so if $P(X)=\sum a_k X^k$ then $\overline{P(t+i)} = \overline{ \sum a_k (t+i)^k } = \sum \overline{a_k (t+i)^k} = \sum a_k \overline{(t+i)^k} $ because $a_k$ are real Am I mistaken ? $\endgroup$
    – W.314
    Nov 19, 2020 at 23:16
  • $\begingroup$ You never claimed that $P(x)$ is a real polynomial (in the sense that its coefficients are real), only that its roots are real. $\endgroup$ Nov 20, 2020 at 7:09
  • $\begingroup$ Well if it has $n$ real roots and it is of degree $n$ I think it is sufficient to conclude that it is a real polynomial don't you agree ? $\endgroup$
    – W.314
    Nov 20, 2020 at 23:49
  • $\begingroup$ If $P(X)=iX$, then all roots of $P(X)$ are real, but $P(X)$ is not a real polynomial, right?! $\endgroup$ Nov 20, 2020 at 23:51

2 Answers 2


We can write $P(x)=\prod_{j=1}^n(x-a_j)$ where $a_j$'s are real roots of $P$. Suppose $Q$ has a non-real root $\alpha+\beta i$. Since its conjugate $\alpha-\beta i$ is also a root of $Q$, we may assume $\beta>0$. Then we have $$\begin{align*} |P(\alpha+\beta i+i)|^2 &=\prod_{j=1}^n \left|(\alpha-a_j)+(\beta+1)i\right|^2 \\&=\prod_{j=1}^n \left[(\alpha-a_j)^2+(\beta+1)^2\right]\\&>\prod_{j=1}^n\left[(\alpha-a_j)^2+(\beta-1)^2\right] \\&=|P(\alpha+\beta i-i)|^2, \end{align*}$$ leading to the contradiction to that $Q(\alpha+\beta i)=P(\alpha+\beta i+i)+P(\alpha+\beta i -i)=0$. Hence all roots of $Q$ should be real.


Hint: If $t$ is real, what is the complex conjugate of $t+i$. What does that tell you about $Q$?

Hint 2: This is a heuristic hint. If you know, or assume, that a polynomial has only real roots, then you can assume the polynomial can be put in the for of a product of monomial factors, each of the form $(x-r_i)$, where r_i is a real root of the polynomial. $$a_n\prod_{i=1}^n(x-r_i)$$

The heuristic hint is this. You will exclude all of the polynomials that have complex roots simply by using that form rather than the form: $$a_0 +\sum_{i=1}^n a_nx^n$$

Hint 3: If $Q(z)= 0$, with $z\in\mathbb{C}$, what can one say about $|P(z+i)|$ as it relates to $|P(z-i)|$?

  • 1
    $\begingroup$ To reinforce this: try a few sample $P$ (e.g. $P(x)=x(x-1)(x+1) = x^3-x$) and see if you notice anything about the resultant $Q$. $\endgroup$ Nov 19, 2020 at 23:15
  • 1
    $\begingroup$ @StevenStadnicki: using the word "resultant" rather than "resulting" in this context may be inviting people to fish for red herrings. $\endgroup$
    – Rob Arthan
    Nov 19, 2020 at 23:24
  • $\begingroup$ @RobArthan I hadn't even considered the terminological overlap, d'oh. Good catch. $\endgroup$ Nov 20, 2020 at 0:41
  • $\begingroup$ Frankly I still don't see how to solve can you give me another hint please ? $\endgroup$
    – W.314
    Nov 20, 2020 at 23:50
  • $\begingroup$ I'm not sure how big of a hint you want. Did you try a few sample polynomials as @StevenStadnicki suggested? $\endgroup$ Nov 21, 2020 at 0:01

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