**Problem:** Let \(P(z) = az^2+ bz+c\), where \(a,b,c\) are complex numbers.

\((a)\) If \(P(z)\) is real for all real \(z\), show that \(a,b,c\) are real numbers.

\((b)\) In addition to \((a)\) above, assume that \(P(z)\) is not real whenever \(z\) is not real. Show that \(a=0\).

**Solution:**

\((a)\) As \(P(z)\) is real for all real \(z\), we have \(P(0)=c\) \(=> c\) is real.

\(P(1) = a+b+c\) is real.

\(P(-1) = a-b+c\) is real.

\(P(1) + P(-1) = 2a+2c\) is real.

As \(c\) is real \(=> a\) is also real.

Similarly as \((a+b+c)\) is real \(=> (a+b+c)-(a+c)\) is also real.

Implying \(b\) is also real.

Thus all \(a,b,c\) are real.

\((b)\)Let us assume that \(a\neq 0\).

Thus the equation can be written as \(P'(z)=z^2 + \frac{b}{a} z + \frac{c}{a} = 0\)

Let \(\alpha\) be a root of the equation. If \(\alpha\) is imaginary that means \(P'(\alpha)\) is imaginary. But \(P'(\alpha)=0\), thus \(\alpha\) is real. Similarly \(\beta\), the other root of the equation, is also real.

Therefore \(\alpha + \beta = -\frac{b}{a}\). \(\cdots (i)\)

Take \(x=\frac{\alpha + \beta}{2} + i\)

Then \(P'(x) = \frac{(\alpha + \beta)^2}{4} + (\alpha + \beta)i -1 + \frac{b}{a}(\frac{\alpha + \beta}{2} + i) + \frac{c}{a}\)

\(=> P'(x) = \frac{(\alpha + \beta)^2}{4} + (\alpha + \beta)i -1 + \frac{b}{a} \frac{\alpha + \beta}{2} +\frac{b}{a} i + \frac{c}{a}\)

Using \((i)\), we get,

\(=> P'(x) = \frac{(\alpha + \beta)^2}{4} – \frac{b}{a} i -1 + \frac{b}{a} \frac{\alpha + \beta}{2} +\frac{b}{a} i + \frac{c}{a}\)

\(=> P'(x) = \frac{(\alpha + \beta)^2}{4} -1 + \frac{b}{a} \frac{\alpha + \beta}{2} + \frac{c}{a}\)

Thus \(P'(x)\) is real even when \(x\) is imaginary. Thus our assumption that \( a \neq 0\) is wrong.

Hence Proved \(a=0\).