It was asked at the [Bulletin of the American Mathematical Society Volume 64, Number 2, 1958](http://www.ams.org/journals/bull/1958-64-02/S0002-9904-1958-10164-0/S0002-9904-1958-10164-0.pdf), as a Research Problem, if a Hurwitz polynomial with real coefficients (i.e. all of its zeros have negative real parts) can be divided into the arithmetic sum of two or three polynomials, each of which has positive coefficients and only nonpositive real roots.


I would like to know if the following problem is known and how it can be solved:



> 
> Can any polynomial with complex coefficients and degree $n$ be divided into the arithmetic sum of three complex polynomials, each of which has degree at most $n$ and only real roots?
> 
> 
> 


Any help would be appreciated.



All polynomials are the sum of three others, each of which has only real roots
To address the original question about the polynomials with *complex* coefficients. 


Given a polynomial $P\in\mathbb C[x]$ of degree $n$, write $P=Q+iR$ with $Q,R\in\mathbb R[x]$, and fix arbitrarily a polynomial $S\in\mathbb R[x]$ of degree $n$ with all roots real and pairwise distinct. As observed in Eremenko's answer, for any $\varepsilon\ne 0$ sufficiently small in absolute value, the polynomials $Q\_\varepsilon:=S+\varepsilon Q$ and $R\_\varepsilon:=S+\varepsilon R$ will also have all their roots real, and then
\begin{align\*}
 P &= Q+iR \\
 &= \varepsilon^{-1}(Q\_\varepsilon-S)+i\varepsilon^{-1}(R\_\varepsilon-S) \\
 &= \varepsilon^{-1}Q\_\varepsilon+i\varepsilon^{-1}R\_\varepsilon-(1+i)\varepsilon^{-1}S
\end{align\*}
is the required representation of $P$ as a sum of three polynomials of degree at most $n$ with all their roots real.


