# A class of (invariant) polynomial equations with real and purely imaginary coefficients

The fundamental theorem of algebra says: If a polynomial equation has complex coefficient(s) it has atleast one complex root. It further follows that if the coefficients are purely real then the equation will have real or complex-conjugate roots.

Following about two decades old complex PT-symmetric quantum mechanics, we can have an interesting class of polynomial equations with real and puely imaginary coefficients which are invariant under changing $$x\to -x$$ and $$i \to -i$$, jointly. For example, see the equations: $$3x^2+2ix+4=0, ~3x^4+4ix^3+x^2+2ix+3=0,~ ix^2+x+i=0, ~ix^4-3ix^2+5x+2i=0$$ etc but avoid trivial cases like: $$ix^2-3ix+2i=0$$ when all coefficients are purely imaginary.

Interestingly, their real roots are found in pairs of $$\pm a$$, and the complex ones are paired as $$\pm b+ic$$, additionally there may be unpaired purely imaginary root as $$id$$, $$ie$$, here all $$a,b,c,d,e$$ are real numbers. Conversely. If we construct a polynomial equation with six roots as $$\pm 1,\pm 1+i, i, 2i$$, we get an invariant polynomial equation as: $$x^6-5ix^4-11x^4+15ix^3+14x^2-10ix-4=0$$

The question is: If $$a,b,c,d,p,q$$ are real and if $$x=p+iq$$ is one root of the equation $$x^4-iax^3+bx^2+icx+d=0~~~~~(1)$$ prove that $$x=-p+iq$$ will also be a root of this interesting (invariant) equation (1).

If $$P(x)=x^4-iax^3+bx^2+icx+d$$ we get that $$Q(x)=P(ix)=x^4-ax^3-bx^2-cx+d$$ has real coefficients and $$-q+ip$$ is a root of $$Q$$ hence $$-q-ip$$ is also a root of $$Q$$ which is equivalent to $$-p+iq$$ root of $$P$$ so we are done!