I need to find a real value for d so that the equation $(2-3i)z^2 -(d-1)z + 4+3i= 0$ has real roots.
I've tried using the discriminant formulae and then I got
$$\frac {(d-1) \pm \sqrt{d^2 -2 d-67 +24i}}{4-6i} $$
Here I got stuck because I get complex numbers.
I appreciate your help.
Let indicate with $w$ the real solution then $w=\bar w$ and consider the conjugate equation
$$(2-3i)w^2 -(d-1)w + 4+3i= (2+3i)\bar w^2 -(d-1)\bar w + 4-3i$$
$$2(w^2-\bar w^2)-3i(w^2+\bar w^2) -(d-1)(w-\bar w) + 6i= 0$$
$$-3i(w^2+\bar w^2) + 6i= 0 \implies w^2+\bar w^2=2\implies w=\pm 1$$
then
Even if you don't have gimusi's insight into considering the conjugate equation, it's possible to "plod" through to a solution.
Let $z = x + yi$. Ultimately, we want to set $y=0$ for the real root(s), but for now, let's do the algebra. $z^2 = x^2 - y^2 + 2xyi$, so the equation becomes:
$$(2-3i)(x^2-y^2+2xyi) - (d-1)(x+yi) + 4+3i = 0$$
$$2(x^2 - y^2) + 4xyi - 3(x^2-y^2)i + 6xy - (d-1)x - (d-1)yi + 4 + 3i = 0$$
And here it should be intuitive that we need to gather terms and set the real and imaginary parts separately to zero.
$$2(x^2 - y^2) + 6xy - (d-1)x + 4 = 0$$
$$4xy - 3(x^2 - y^2) - (d-1)y + 3 = 0$$
Set $y = 0$ in both equations. The second immediately yields: $x = \pm 1$. Substituting these two possible values for $x$ into the first equation yields the two possible values $d = -5, 7$.
If $z\in \mathbb{R}$ and $$(2-3i) z^2 -(d-1)z + 4+ 3i = 0$$
then $2z^2-(d-1)z+4=0$ and $-3z^2 +3=0$ (both real and imaginary part of left side must be $0$) so $z=\pm 1$ and $$d={2z^2+z+4\over z}=...$$
