TITLE: How to show $ a,b,c \in \mathbb R , z \in \mathbb C , az^2 + bz + c = 0 \iff a\bar{z}^2 + b\bar{z} + c = 0$?
QUESTION [5 upvotes]: I am completely stumped by with this problem.

If a, b and c are real numbers and z is a complex number of the form $x + yi$, prove that $$az^2 + bz + c = 0 \iff a(z^*)^2 + b(z^*) + c = 0$$

I tried substituting $z = x + yi$ and for $z^*= x - yi$, but that gives a bigger equation with multiple variables. How should I proceed with this? Thanks for your help.

REPLY [3 votes]: Here is a solution along the lines you took:
$a(z^*)^2 + b(z^*) + c = a(x-yi)^2 + b(x-yi) + c$
$\quad=a(x^2-2xyi-y^2) + b(x-yi) + c $
$\quad= (a(x^2-y^2)+bx+c) - (2axy+by)i$
$\quad=(az^2 + bz + c)^*$
Since $w=0$ iff $w^*=0$, you get your result.