Cube root of numbers such as $2+11i$ How can I find powers and roots of complex numbers with ugly argument such as cube root of 2+11i? I saw Find the solutions to $z^3 = 2 + 11i$., but the answer includes some guessing and I would like to have some algorithm for such a tasks, which I can program. I would also like to solve it symbolically and thus precisely. Of course I know the formula: $a^b=e^{b \ln(a)}$ But when I try to use it:
$\sqrt[3]{2+11i}=(2+11i)^\frac{1}{3}=e^{\frac{1}{3} \ln(2+11i)}=e^{\frac{1}{3} \big(\ln(\sqrt{125})+i \arctan(\frac{11}{2}) \big)}=$
$=\sqrt{5} \big(\cos(\frac{1}{3} \arctan(\frac{11}{2}))+i \sin (\frac{1}{3} \arctan(\frac{11}{2})) \big)$
And I cannot solve this further symbolically, of course when I solve it in floating point numbers I get the right solution $2+i$.

Note that $\tan 3\alpha=\dfrac {t(3-t^2)}{1-3t^2}$, where $t=\tan\alpha$.
Putting $\theta=3\alpha$ where $\tan\theta=\tan3\alpha=\dfrac {11}2$, we have
$$\begin{align}
\frac {11}2&=\frac {t(3-t^2)}{1-3t^2}\\
11(1-3t^2)&=2t(3-t^2)\\
2t^3-33t^2-6t+11&=0\end{align}$$
By inspection, putting $t=\frac 12$ gives $\text{LHS}=0=\text {RHS}$,  hence $t=\tan\alpha=\tan\frac \theta 3=\frac 12$ is a solution. 
$$\large\begin{align}
\left[re^{i(2n\pi+\theta)}\right]^{\frac 13}&=r^{\frac 13}e^{i\left(\frac {2n\pi+\theta}3\right)}\\&=\big(\sqrt{125}\big)^\frac 13e^{i\left(\frac {2n\pi}3+\frac\theta 3\right)}
&&\scriptsize{\theta=\arctan(\frac {11}2)}\\
&=\sqrt{5}e^{i\left(\frac {2n\pi}3+\frac\theta 3\right)}\\
&=\sqrt{5}e^{i\left(\frac {2n\pi}3+\arctan\frac 12\right)}\end{align}$$
Hence the principal solution  ($n=0)$ is 
$$\begin{align}
\large\sqrt{5}e^{i\cdot\arctan\frac 12}
&=\sqrt{5} \left(\cos (\arctan \left(\tfrac 12\right)+i\sin(\arctan\left(\tfrac 12\right)\right)\\
&=\sqrt{5}\left(\frac 2{\sqrt5}+i\frac 1{\sqrt{5}}\right)\\
&=2+i\end{align}$$
