Intervals for coefficients, complex equation

I'm not sure what is being asked for. Perhaps that is why you haven't any response.

Making assumptions if the roots of p(z) are imaginary they are complex conjugates

Z0=-(a/2)+[4b-a^2]^(1/2)i
&
Z1=-(a/2)-[4b-a^2]^(1/2)i

to put this on the circle cos(th)^2+sin(th)^2=1
then you need to fine Zr=[(a/2)^2+{[4b-a^2]^(1/2)/2}^2]^(1/2)=b^(1/2)
where Zr=b^(1/2)<1
so
Z0=Zr{-(a/2)/Zr+[4b-a^2]^(1/2)/Zri}
&
Z1=Zr{-(a/2)/Zr-[4b-a^2]^(1/2)/Zri}
which is the same as
Z0=Zr{cos(th)+sin(th)i}
&
Z1=Zr{cos(th)-sin(th)i}
so
cos(th)= (a/2)/Zr
&
sin(th)=[4b-a^2]^(1/2)/Zr
so
th=arcsin([4b-a^2]^(1/2)/Zr)
& the interval between those conjugates is twice that angle.

Rap

Gee rap-

I really do feel in the presence of a near genius.I'm not qualified to go further.

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I notice I am not the only one you insult with impunity Spendius.

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I meant it as a compliment you silly old clunker.