Find $a$ and $b$ such that $\Re(az^2+bz)\leq \Im(az^2+bz)$

Question

Find the complex numbers $a$ and $b$ such that

$$\Re(az^2+bz)\leq \Im(az^2+bz)$$

for any complex number $z \in \mathbb{C}$.

I let $a=x_1+iy_1, b=x_2+iy_2$ and I tried to set values for $z$:

$z=1 \implies x_1+x_2\leq y_1+y_2$

$z=-1 \implies x_1-x_2\leq y_1-y_2$

If I sum, I find $x_1 \leq y_1$ and I have no idea how to follow up this.

Answer 1

HINT:

If $a$, $b$ are not both zero then for every $c\in \mathbb{C}$ there exist $z\in \mathbb{C}$ so that $$a z^2 + b z =c$$ (every nonconstant polynomial function is surjective).

So only $a=b=0$ work.