jmc

Let $r=|a|=|b|=|a+b|.$ Then $$a\overline{a}=b\overline{b}=r^2,$$so $\overline{a}=\frac{r^2}{a}$ and $\overline{b}=\frac{r^2}{b}.$

Also, $(a+b)(\overline{a+b})=r^2.$ Then $(a+b)(\overline{a}+\overline{b})=r^2,$ so $$(a+b)\left(\frac{r^2}{a}+\frac{r^2}{b}\right)=r^2.$$Then $$(a+b)\left(\frac{1}{a}+\frac{1}{b}\right)=1,$$which expands as $$1+\frac{a}{b}+\frac{b}{a}+1=1,$$so $$\frac{a}{b}+\frac{b}{a}=-1.$$Let $z=\frac{a}{b}.$ Then $z+\frac{1}{z}=-1,$ so $z^2+1=-z,$ or $$z^2+z+1=0.$$By Vieta's formulas, the sum of the roots is $\boxed{-1}.$