Let P(z)=az2+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.

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membatas0v2v · Accepted Answer

Suppose a≠0, the we show that p(z) can be real for some complex z∈C∖R. Let c=α+βi. Then p(z)=az2+bz+(α+βi). This is real if the imaginary parts all cancel, that is, if there is some z∈C∖R so that az2+bz+βi=0. But this is just a quadratic in z so there are two solutions unless b2−4aβ=0, then z=−b2a is complex unless b is some multiple of a. If there are two solutions, then just convince yourself that you can pick a, b, β so that both solutions are complex.

Let \(\displaystyle{P}{\left({z}\right)}={a}{z}^{{2}}+{b}{z}+{c}\), where a,b,c are complex numbers. 1)

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