How do I find a constant for a polynomial so its roots are reflective around a linear function? How can I find all complex numbers $w$ so that the roots of the following polynomial are reflected around a linear function $f(x)$ $$p(q) = q^2-4q+w = 0$$
If I want to find all the complex numbers $w$, for which the roots of the polynomials are reflective around, say $f(x) = -x+2$, how would I go about doing that?
Mark my words! The text below is a bit confusing, just showing my work if it lights any spark in someone's mind, hopefully it does! I really want to solve this problem, seems like something that would be very useful to know how to solve.

I have been playing around and noticed some few things.

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*If we have a function $g(x)$ which is perpendicular to $f(x)$ then all roots in a polynomial must either come in pair as points on $g(x)$ line OR a root will be sitting on the $f(x)$ line [double root?]; this is the requirement for the condition in the title of this question


*This means in my case that all roots will be in the form of $\alpha=r\cdot e^{i\cdot(\pi/4+k\pi)}+ z$ where $z$ is the point at which $f(x) = g(x)$. And $r=\sqrt{(\alpha_1-\Re z)^2+(\alpha_1-\Im z)^2}$, where $\alpha_1$ is a root.
The reason behind this is that the perpendicular function $g$ would have a negated slope from $f$ which in my case had the slope of $-1\cdot x\implies g(x) = x + m$ where $m$ would be chosen according to a root found. Drawing this, one easily sees the roots must be found at $\theta=\pi/4\ or\ -3\pi/4$ with mid-point in $z$, and a radius of $r$.


*I have also tried using Vietas formulas with no luck as Vietas formulas requires all coefficients to be known for a full solution. (Obviously)
Edit 1. I might add!
I'm still looking for a complete solution for all possible $w$ where the roots of $p$ are reflective around $f(x) = -x+2$
 A: Using the Q.F., the roots of
$$p(q) = q^2 + B q + C$$
are
$$r_{\pm} = \frac{1}{2}\left(-B \pm \sqrt{B^2 - 4 C}\right)$$
(here $\sqrt{\cdot}$ is any branch).
If $p$ has a double root, which the above shows happens iff $B^2 = 4C$, in which case the double root is at $-\frac{B}{2}$, then any reflection through a line containing $2$ maps each root to the other. So, henceforce suppose $p$ has two distinct roots.
Now, the reflection that exchanges $r_{\pm}$ is exactly a reflection across the perpendicular bisector of the line segment with vertices $r_{\pm}$: By construction this line contains the midpoint $\frac{1}{2}(r_+ + r_-) = -\frac{B}{2}$, and is orthogonal to the vector $\sqrt{B^2 - 4C}$, so it equation is
$$\sqrt{B^2 - 4C} \cdot \left(z + \frac{B}{2}\right) = 0,$$
where $\cdot$ denotes the dot product $a \cdot b := \Re(\bar{a} b).$
On the other hand, we can write the line $ax + by = c$ through $-\frac{B}{2}$ in complex form as
$$(a + bi) \cdot \left(z + \frac{B}{2}\right) = 0.$$ So, reflection across this line exchanges the roots of $p$ if
$$a + bi = \sqrt{B^2 - 4C},$$
that is, if
$$(a^2 - b^2) + 2abi = B^2 - 4C.$$
In your case, $B = -4$, and $C = w$, and we can take $a = b = 1$, and substituting and rearranging gives
$$w = \frac{1}{4}(16 - 2i) = 4 - \frac{1}{2}i.$$ Of course, we can just ask well take $a = b = \lambda$ for any nonzero (real) $\lambda$, which leads to the general solution
$$w = 4 - \frac{1}{2}\lambda^2 i, \qquad \lambda \neq 0,$$
which we can suggestively write as
$$w = 4 + \mu i, \qquad \mu < 0.$$
