QUESTION: Let $P(z)=az^2+bz+c$ , where $a,b,c$ are complex numbers.
1) If $P(z) $ is real for all real $z$ then show that $a,b,c$ are real numbers.
2) In addition to (1) above, assume that $P(z)$ is not real whenever $z$ is not real. Show that $a=0$.
MY ANSWER: (1) was easy. Clearly, $P(0)=c$. Therefore, $c$ is real. Similarly, by doing $P(1)+P(-1)$ we can show $a$ is real. And it also follows that $b$ is real.
But for (2) I am a bit confused. I assumed $z=x+yi$, where $i=\sqrt{-1}$. But, somehow I cannot come to a satisfactory conclusion. Can anyone help me out to prove no.(2)? Thank you.
Suppose $a \neq 0$, the we show that $p(z)$ can be real for some complex $z \in \mathbb{C} \setminus \mathbb{R}$. Let $c = \alpha + \beta i$. Then $p(z) = az^2 + bz + ( \alpha + \beta i)$. This is real if the imaginary parts all cancel, that is, if there is some $z \in \mathbb{C} \setminus \mathbb{R}$ so that $az^2 + bz + \beta i = 0$. But this is just a quadratic in $z$ so there are two solutions unless $b^2 - 4a\beta = 0$. If $b^2 - 4a\beta = 0$, then $z = -b/2a$ is complex unless $b$ is some multiple of $a$. If there are two solutions, then just convince yourself that you can pick $a,b,\beta$ so that both solutions are complex.