jmc

We can write $|z{|}^2=z\overline{z},$ so the equation becomes $$z^3+z\overline{z}+z=0.$$Since $a$ and $b$ are positive, $z=a+bi$ is nonzero. Thus, we can divide both sides of the equation above by $z,$ which gives us $$z^2+\overline{z}+1=0.$$Then $(a+bi{)}^2+\overline{a+bi}+1=0,$ or $$a^2+2abi-b^2+a-bi+1=0.$$Equating real and imaginary parts, we get $$\begin{array}{rl}{a}^2-b^2+a+1& =0,\\ 2ab-b& =0.\end{array}$$From the second equation, $b(2a-1)=0.$ Since $b$ is positive, $2a-1=0,$ so $a=\frac{1}{2}.$ Then from the first equation, $$b^2=a^2+a+1=\frac{7}{4}.$$Since $b$ is positive, $b=\frac{\sqrt{7}}{2}.$ Thus, $(a,b)=\boxed{\left(\frac{1}{2},\frac{\sqrt{7}}{2}\right)}.$