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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 $2+i$.

José Carlos Santos
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Martin Schmied
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3 Answers3

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There is, really, no algebraic formula for extracting the cube roots of complex numbers. You can try rendering the equations into real variables as the OP did. You end up with a cubic equation that can be solved in several ways -- but none that lends itself to a useful algebraic formula.

You can use the Rational Root Theorem to identify good candidates for guessing, but there is a chance that none of these educated guesses works.

Or you can try the general formula from Cardano only to find that you get back the complex radical you intended to solve.

Or you can "cheat", going beyond algebraic methods, by going to trigonometric functions, which is basically the polar form solution. The trigonometric solution, of course, appears in a form that offers no clue it might be a rational root.

Oscar Lanzi
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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.

José Carlos Santos
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  • Thanks for answer. But is it not possible to somehow continue in calculation using the formula in my post? I discovered, that WolframAlpha knows, that $\frac{1}{3} arctan(\frac{11}{2})=arctan(\frac{1}{2}) $ and it appears to have calculated it symbolically. How is it possible? – Martin Schmied Jun 30 '18 at 10:50
  • We have$$\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right).$$Therefore\begin{align}3\arctan\left(\frac12\right)&=\arctan\left(\frac12\right)+\arctan\left(\frac12\right)+\arctan\left(\frac12\right)\&=\arctan\left(\frac43\right)+\arctan\left(\frac12\right)\&=\arctan\left(\frac{11}2\right).\end{align} – José Carlos Santos Jun 30 '18 at 11:36
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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 $$\begin{align} \large\sqrt{5}e^{i\cdot\arctan\frac 12} &=\sqrt{5} \left(\cos (\arctan \left(\tfrac 12\right)+i\sin(\arctan\left(\tfrac 12\right)\right)\\ &=\sqrt{5}\left(\frac 2{\sqrt5}+i\frac 1{\sqrt{5}}\right)\\ &=2+i\end{align}$$

Hypergeometricx
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