# complex polynomial has zeroes only in the upper half plane

Let $f(z)=z^{n}+a_{1}z^{n-1}+...+a_{n}$ be a polynomial with complex coefficients and suppose it has $n$ zeros in the upper half plane, that is $\operatorname{Im} z>0$, and let $\alpha_ {k}$ be the real part of $a_{k}$. Show that $\alpha(x)=x^{n}+\alpha_{1}x^{n-1}+...+\alpha_{n}$ has $n$ real distinct roots.

• I assume $f$ has distinct zeroes for otherwise the statement is false – ArtW May 22 '18 at 9:42
• @BillO'Haran observed it in some easy case,n=1,2,but i found nothing – user561425 May 22 '18 at 9:45
• Are you able to proceed from my answer? – Bill O'Haran May 22 '18 at 10:05
• @BillO'Haran thanks for your hint! – user561425 May 22 '18 at 10:13
• You're welcome :) – Bill O'Haran May 22 '18 at 10:13

Hint:

Your problem comes down to showing that $P + \overline{P}$ has $n$ real roots. Then write $P = \prod_{k=1}^n (z-z_k)$ and notice how $|z-z_k|<|z-\overline{z_k}|$ if Im $z>0$.

Whole solution:

$z$ is a root of $P + \overline{P}$ iff: $$\prod_{k=1}^n (z-z_k)= - \prod_{k=1}^n (z-\overline{z_k})$$ If Im $z>0$, then: $$\forall k\in \{1,\dots,n\}, |z-z_k|<|z-\overline{z_k}|$$ And: $$\left| \prod_{k=1}^n (z-z_k)\right|< \left|\prod_{k=1}^n (z-\overline{z_k})\right|$$ Thus, $z$ is not a root of $P+\overline{P}$ and neither is $\overline{z}$ (because $P+\overline{P}$ is a real polynomial).

• +1, great idea! How would you show that the roots are distinct? They are even if $P(x)=(x-i)^n$ . – Orest Bucicovschi May 22 '18 at 11:42
• @orangeskid the distinct is another challenge，we need claim that the degree of f is even. – user561425 May 22 '18 at 12:24

the distinct comes from the following : Since $$\alpha(x)=\prod_{k=1}^{n}(x-\alpha_k)$$ $$\alpha'(x)=\alpha(x)\sum_{k=1}^{n}\frac{1}{x-\alpha_k}$$