jmc

We start with small cases. For $n=1,$ the equation becomes $$a+bi=a-bi,$$so $2bi=0,$ which means $b=0.$ This is not possible, because $b$ is positive.

For $n=2,$ the equation becomes $$a^2+2abi-b^2=a^2-2abi-b^2=0,$$so $4abi=0,$ which means $ab=0.$ Again, this is not possible, because both $a$ and $b$ are positive.

For $n=3,$ the equation becomes $$a^3+3a^{2}bi+3a{b}^2{i}^2+b^3{i}^3=a^3-3a^{2}bi+3a{b}^2{i}^2-b^3{i}^3,$$so $6a^{2}bi+2b^3{i}^3=0,$ or $6a^{2}bi-2b^{3}i=0.$ Then $$2bi(3a^2-b^2)=0.$$Since $b$ is positive, $3a^2=b^2.$ Then $a\sqrt{3}=b,$ so $\frac{b}{a}=\boxed{\sqrt{3}}.$