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\begin{align}(a+bi)^3=2+11i&\iff\left\{\begin{array}{l}a^3-3ab^2=2\\3a^2b-b^3=11.\end{array}\right.\\&\iff\left\{\begin{array}{l}a(a^2-3b^2)=2\\b(3a^2-b^2)=11\end{array}\right.\end{align}Perhaps that there are no integer solutions, but if there are, then you can use the fact that both $2$ and $11$ are prime numbers. The only way of expressing $2$ as a product of two integers is when one of the factors is $\pm2$ and the other one is $\pm1$. So, can we take $a=2$ and $a^2-3b^2=1$. Yes, we can! Just take $b=\pm1$. And it turns out that $a=2$ and $b=1$ are also a solution of the second equation.
