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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).

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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!

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