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.
Best Answer
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.