Avadhoot's Algebra problem

Let $f(m+ni)=z$ where $(m,n)\in\mathbb{R}$ and $n≠0$ . We then have $\Im{(z)}$ = $2amn+bn$ = $n(2am+b)$ According to conditions of problem, $\Im{(z)}≠0$ Thus, $n(2am+b)≠0$ Since $n≠0$ , we must have $2am+b≠0$ , but if $m=\frac{-b}{2a}$ then $2am+b=0$ . This can be avoided iff $a=0$ .

Let's prove first that a=0. If a≠0, then we can prove that for some non real z we'll have az²+bz as a real number. Now let that real number be k. So now we want az²+bz=k where a,b and k are real numbers and a≠0. We can always find such k so that the above equation has non real roots. So the expression ax²+bx+c will attain a real value for at least one non real x if a≠0 . So a=0. b and c are not necessarily 0 as we take an example 2x+1 which satisfies all the given conditions.

It is not necessary that $c = 0$ , as $(ax^{2} + bx) \in \mathbb{R} \Rightarrow (ax^{2} + bx + c) \in \mathbb{R}$ and $(ax^{2} + bx) \in \mathbb{C} \setminus \mathbb{R} \Rightarrow (ax^{2} + bx + c) \in \mathbb{C} \setminus \mathbb{R}$ ( $\forall c \in \mathbb{R}$ )

So the first and the third option are ruled out.

If $b=0$ is necessary (the second option), it is clear that it isn't sufficient in that it necessarily implies $a = 0$ (otherwise: $a*(i^{2})+c = c - a \in \mathbb{R}$ ).

But as only one option is correct this leaves us with the fourth option, that $a = 0$ .