# Find all complex numbers $z$ such that given quadratic has real solution

I have a question about this question. Find all complex numbers $z$ such that the equation $$t^2 + [(z+\overline z)-i(z-\overline z)]t + 2z\overline z\ =\ 0$$ has a real solution $t$.

Attempt at a solution

The discriminant is

$[(z+\overline z) - i(z-\overline z)]^2 - 4(2z\overline z)$
$=\ (z+\overline z)^2 - 2i(z+\overline z)(z-\overline z) + [i(z-\overline z)]^2 -8z\overline z$
$=\ (z^2+2z\overline z+\overline z^2) -2i(z^2-\overline z^2) - (z^2-2z\overline z+\overline z^2)-8z\overline z$
$=\ -4z\overline z - 2iz^2 + 2i\overline z^2$

For real solutions, the discriminant must be non-negative. But $z$ is a complex number; how can complex numbers be positive or negative? This is what I don't understand.

Would appreciate any help. Thanks.

• Your title needs improvement... – A. Chu Jan 7 '17 at 10:30

with $$z=x+iy$$ we get $$\bar z=x-iy$$ thus our equation is given by $$t^2+2(x+y)t+2(x^2+y^2)=0$$ can you proceed?
Hint: Letting $z=a+ib$ reduces the quadratic equation to one with only real coefficients.
Just set $z=u+iv$. The discriminant becomes $$\Delta=4[(u+v)^2-2(u^2+v^2)]=-4(u-v)^2.$$ Hence the condition is $\;u=v,\;$ or $\;\arg z\equiv\dfrac\pi4\mod\pi$.
• Why isn't the discriminant $4[(u+v)^2-2(u^2+v^2)]$? – Vera Jan 7 '17 at 10:34