${z_n}$ is a geometric sequence, $z_1=1, z_2=a+bi, z_3=b+ai$, $a$ and $b$ are real numbers, $a>0$. What is the smallest $n$ such that $z_1+z_2+...+z_n=0$?
So far I found out that $r=\frac{1}{2}+\frac{\sqrt{3}}{2}i$, but I don't know how to find the smallest $n$ such that $z_1+z_2+...+z_n=0$. I tried using the geometric series formula and got:
$\frac{1-(\frac{1}{2}+\frac{\sqrt{3}}{2}i)^n}{1-(\frac{1}{2}+\frac{\sqrt{3}}{2}i)}=0$, and from here multiplied both sides by ${1-(\frac{1}{2}+\frac{\sqrt{3}}{2}i)}$ to get $1-(\frac{1}{2}+\frac{\sqrt{3}}{2}i)^n=0$, and then moved one to the other side, got rid of the negatives, and took the n root of both sides. But, then I got: $\frac{1}{2}=\frac{\sqrt{3}}{2}i$. So, I don't know if I messed up or if there's no solution.
Note: I think you have calculated $r$ incorrectly. Based on what you have $$z_3=rz_2=r^2 \implies b+ai=(a+ib)^2.$$ This gives: \begin{align*} b&=a^2-b^2\\ a&=2ab. \end{align*} Solving this gives us either the zero solution or $a=\pm \frac{\sqrt{3}}{2}$ and $b=\frac{1}{2}$. Thus $\color{red}{r=\frac{\pm\sqrt{3}+i}{2}=e^{i\frac{\pi}{6}} \text{ or } e^{i\frac{5\pi}{6}}}$
For the first value $r=e^{i\frac{\pi}{6}}$. So, \begin{align*} z_1+z_2+\dotsb+z_n&=1+r+r^2+\dotsb+r^n\\ &=\frac{1-r^{n}}{1-r}. \end{align*} For this to be zero, we need $$r^{n}=1 \implies e^{i\frac{n\pi}{6}}=1.$$