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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.

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

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    $\begingroup$ +1, great idea! How would you show that the roots are distinct? They are even if $P(x)=(x-i)^n$ . $\endgroup$ – Orest Bucicovschi May 22 '18 at 11:42
  • $\begingroup$ @orangeskid the distinct is another challenge,we need claim that the degree of f is even. $\endgroup$ – user561425 May 22 '18 at 12:24
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the distinct comes from the following : Since $$\alpha(x)=\prod_{k=1}^{n}(x-\alpha_k)$$ $$\alpha'(x)=\alpha(x)\sum_{k=1}^{n}\frac{1}{x-\alpha_k}$$

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