jmc

Let $z=x+yi,$ where $x$ and $y$ are real numbers. Then both $$z+\overline{z}=(x+yi)+(x-yi)=2x,$$and $$z\overline{z}=(x+yi)(x-yi)=x^2+y^2$$are real numbers. Therefore, by Vieta's formulas, all the coefficients must be real numbers. Then $(a,b)=\boxed{(0,0)}.$