# Strange functioning of RegionPlot with Reduce's output

I have defined this two quantities:

f = (-b + Sqrt[b^2 + 4 a])/(2 a)
g = (-b - Sqrt[b^2 + 4 a])/(2 a)


and I want to solve this system in function of $a$, for $f,g$ complex: $$\begin{cases} |f(a,b)|>1 \\ |g(a,b)|>1 \\ Im(a)=Im(b)=0 \end{cases}$$

So I use:

Reduce[Abs[f] > 1 && Abs[g] > 1 && Im[a] == 0 &&  Im[b] == 0, a, Complexes]


getting the solution I seached. Now, I want to split the regions where the square root in the equation of $f,g$ is positive or negative. If I use:

Reduce[Abs[f] > 1 && Abs[g] > 1 && Im[a] == 0 &&   Im[b] == 0 && (Re[b^2 + 4 a] >= 0 ||
Re[b^2 + 4 a] < 0) , a, Complexes]


I got the same result as before, and this is true. My ploblem is that if a solve separately

Reduce[Abs[f] > 1 && Abs[g] > 1 && Im[a] == 0 && Im[b] == 0 &&   Re[b^2 + 4 a] >= 0 , a, Complexes]
Reduce[Abs[f] > 1 && Abs[g] > 1 && Im[a] == 0 && Im[b] == 0 &&   Re[b^2 + 4 a] < 0 , a, Complexes]


and I plot the results using RegionPlot a small area disapears. How can I solve this problem?

• @Öskå Thank you for your answer. I added MaxRecursion to the options of RegionPlot and I fixed the problem. Can you explain me why without it RegionPlot does not work? Oct 8 '14 at 14:22
• See MaxRecursion for the details :)
– Öskå
Oct 8 '14 at 14:23

You need to increase the MaxRecursion:

f = (-b + Sqrt[b^2 + 4 a])/(2 a);
g = (-b - Sqrt[b^2 + 4 a])/(2 a);
RegionPlot[{
Reduce[Abs[f] > 1 && Abs[g] > 1 && Im[a] == 0 &&
Im[b] == 0 && Re[b^2 + 4 a] >= 0, a, Complexes],
Reduce[Abs[f] > 1 && Abs[g] > 1 && Im[a] == 0 &&
Im[b] == 0 && Re[b^2 + 4 a] < 0, a, Complexes]}, {a, -2, 2}, {b, -3, 3},
MaxRecursion -> 7, Evaluated -> True] • What is the option Evaluated -> True for RegionPlot? I was not able to find it in the documentation. Jul 20 '17 at 18:53
• @TaikiBessho, Try it without, and you will see the difference :) You can always look for the Evaluated documentation.
– Öskå
Jul 20 '17 at 20:28