complex polynomial has zeroes only in the upper half plane

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 $z>0$.

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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).

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