Location of zeros of the "real part" of a polynomial

Question

I came across the following question on an old qualifying exam:

Let $p$ be a polynomial, all of whose zeros lie in the lower half plane $\lbrace z : \text{Im}(z) < 0 \rbrace$. Let $a$ and $b$ be the unique pair of polynomials with real coefficients such that $p(z) = a(z) + ib(z)$. Prove that $a$ and $b$ have only real zeros.

I tried to think of a creative way of using the argument principle to approach this, but my attempts so far have been unsuccessful. Does anyone have any insight into this problem? A solution or hint would be appreciated.

Answer 1

This is very simple if you think about how $a$ and $b$ are related to the factored form of $p$. Note first that for the result to be true, you must assume that $p$ is not constant (if $p$ is a real or purely imaginary constant then one of $a$ and $b$ is identically zero). Let $r_1,\dots,r_n$ be the roots of $p$, so $p(z)=c\prod(z-r_k)$ for some nonzero constant $c$. Let $q(z)=\overline{c}\prod (z-\overline{r_k})$, and observe that the coefficients of $q$ are conjugate to the coefficients of $p$ so $a(z)=\frac{p(z)+q(z)}{2}$ and $b(z)=\frac{p(z)-q(z)}{2i}$.

Now observe that if $\operatorname{Im}(z)>0$ then $|z-r_k|>|z-\overline{r_k}|$ for each $k$ since $\operatorname{Im}(r_k)<0$. Thus $|p(z)|>|q(z)|$ (here we use the assumption that $p$ is not constant). In particular, $p(z)$ cannot be equal to $\pm q(z)$, so $a(z)$ and $b(z)$ must be nonzero. Similarly, if $\operatorname{Im}(z)<0$, then $|p(z)|<|q(z)|$ so $a(z)$ and $b(z)$ must be nonzero. Thus any zero of $a$ or $b$ must be real.

Answer 2

Assume $p\in \Bbb{C}[z]$ is monic, all its zeros are in the lower half-plane and $p(z)=a(z)+ib(z),a,b\in \Bbb{R}[z]$.

• For $r> 0$ let $p_r(z)=a(z)+rib(z)$

  For $x$ real, $p_r(x)=0 \implies \Re(p_r(x))=\Im(p_r(x))=0\implies p(x)=0$ which is a contradiction.

• Take $T$ large enough so that $p_r(z)=0\implies |z|<T$. Let $U = \{ z,|z|<2T,\Im(z)< 0\}$. We get that

$$\frac1{2i\pi }\int_{\partial U} \frac{p_r'(z)}{p_r(z)}dz = \# \{ z\in U, p_r(z)=0\}$$

is continuous in $r$, ie. all the zeros of $p_r$ are in $U$.

An arbitrary small perturbation of the coefficients of $p_0$ move its zeros in $U$: thus all the zeros of $p_0$ are in $\overline{U}$. Since $p_0$ has real coefficients, a non-real zero would give a pair of zeros whose one is outside $\overline{U}$, and hence all the zeros of $p_0$ are real.

Answer 3

An easy way to see this is to notice that:

$2i(a(\bar z)b(z)-a(z)b\bar(z))=|p(z)|^2-|p(\bar z)|^2$ and the geometrically obvious fact that if $\Im z >0$, $|p(z)| > |p(\bar z)|$ using that $|z-w| > |\bar z -w|$, when $\Im z>0$ and $\Im w <0$, the roots of $p$ satisfy $\Im w <0$ and the decomposition in factors of $p$.

But $a(z)=0$ implies $a(\bar z)=0$ (similarly for $b$) and we can choose such a $z$ s.t. $\Im z >0$ if somehow there is a non-real zero of $a$ (or $b$) and that leads to a contradiction since the RHS above is strictly positive!

(as an aside this is an easy part of the famous Theorem of Hermite-Biehler which states that $p$ as above has all the roots in the half-plane $\Im z<0$ or all the roots in the half-plane $\Im z >0$ if and only if $a,b$ have real and strictly interlacing zeroes - this means the obvious, ie the roots of $a,b$ strictly alternate on the real axis, so in particular their degrees can differ by at most $1$)