Is this correct?
$a^3 =r^3e^{i3\theta}= 5\sqrt{5}e^{i\arctan(11/2)}$ $$\implies r=\sqrt{5}, 3\theta = \arctan(11/2)+2\pi n,n\in\Bbb Z$$ $$\theta = \frac{\arctan(11/2)+2\pi n}{3}$$ $$\theta = \frac{\arctan(11/2)}{3},\frac{\arctan(11/2)-2\pi}{3}, \frac{\arctan(11/2)-4\pi }{3}$$
So $a=\sqrt{5}e^{i\frac{\arctan(11/2)}{3}}, \sqrt{5}e^{i\frac{\arctan(11/2)-2\pi}{3}},\sqrt{5}e^{i \frac{\arctan(11/2)-4\pi }{3}}$
Is there a simplification I am missing?
If you are looking to simplify the angles, then you can use $\frac 13\arctan(\frac{11}{2})=\arctan \frac 12$
You can find this by using $$\tan3\theta=\frac{3t-t^3}{1-3t^2}=\frac{11}{2},$$ where $t=\tan\theta$ $$\Rightarrow2t^3-33t^2-6t+11=0$$ $$\Rightarrow(2t-1)(t^2-16t-11)=0$$ from which the acute angled solution for $\theta$ which is less than $\arctan\frac{11}{2}$ is given by $t=\frac 12$
As an added bonus, the roots of the quadratic factor give the other angles namely $\arctan(8\pm5\sqrt{3})$
