Solving a cube root with complex numbers

Question

A book I own gives me this equation: $\sqrt[3]{\sqrt{-121} + 2}$ and says it can easily be simplified to: $\sqrt{-1} + 2$. I have asked my mathematics teacher and he was not able to figure it out "easily" so I was wondering how do you simplify this?

Answer 1

I would certainly not say that there is such an easy simplification.

Note that you're after numbers $a$ and $b$ such that$$\left(a+b\sqrt{-1}\right)^3=2+\sqrt{-12}=2+11\sqrt{-1}.\tag1$$But$$\left(a+b\sqrt{-1}\right)^3=a^3+3a^2b\sqrt{-1}-3ab^2-b^3\sqrt{-1}.$$Therefore, $(1)$ is equivalent to$$\left\{\begin{array}{l}a^3-3ab^2=2\\3a^2b-b^3=11.\end{array}\right.$$Now, if there are integer solutions of this system, then, since the first equation is equivalent to $a(a^2-3b^2)=2$, you don't have plenty of choices. One of them is $a=2$ and $b=1$ and it turns out that it is also a solution of the second equation. So, yes, $\left(2+\sqrt{-1}\right)^3=2+11\sqrt{-1}$.

Answer 2

First, recognize that $\sqrt{-121}=11i$, so under the cube root, we have $2+11i$. This can be rewritten to $$\sqrt{125}\left(\frac{2}{\sqrt{125}}+\frac{11}{\sqrt{125}}i\right).$$ Using Euler's formula, this is $$\sqrt{125}e^{i\theta}$$ for some $\theta\in(0,\frac{\pi}{2})$ (we can state $\theta$ exactly using inverse trig. functions). This latter expression is under the cube root, so once we take the cube root and noting the $\sqrt[3]{125}=5$, this becomes $$\sqrt{5}e^{i\theta/3}.$$Applying Euler's formula again, this is $$\sqrt{5}\left(\cos(\theta/3)+i\sin(\theta/3)\right).$$

Up to this point, I would consider our work "easy" (depending on the context the problem is presented in within the book) because it doesn't involve any particularly deep thoughts. After this, one must know or derive the formula to $\sin(x/3)$ and $\cos(x/3)$ (which can be derived from the knowledge of cubic equations, e.g., here). Using the formulas for $\sin(\theta/3)$ and $\cos(\theta/3)$, it just becomes a matter of plugging in values and calculating.

Answer 3

$\newcommand{\i}{\mathrm{i}}$

Let $N(x+\i y)=x^2+y^2\text{.}$ If $a+b\i=(c+d\i)^3$ for integer $a$, $b$, $c$, $d$, then $N(a+b\i)=N(c+d\i)^3$, so $N(a+b\i)$ must be a cube of an integer. In our case, $N(2+11\i)=125$, which is the cube of $5$. Therefore it suffices check for and enumerate the solutions to $$c^2+d^2=5$$ Now, it is well known that $$c^2+d^2=p$$ has a solution for prime $p>2$ iff $p\equiv 1\pmod 4$. This is true in our case, so $c^2+d^2=5$ has integer solutions. By guess-and-check, $$c+d\i=2+\i$$ is a solution; the full set of solutions is given by $\i^k(2+\i)$, $k=0,1,2,3$, because if $N(x)=N(y)$ then $N(x/y)=1$ and $N(z)=1\Rightarrow z\in\{1,\i,-1,-\i\}$. Of these, only $2+\i$ cubes to $2+11\i$.

Answer 4

After Carlos,

as $ab\ne0,$

set $a=mb$

$$\dfrac2{11}=\dfrac{m^3-3m}{3m^2-1}$$

$11m^3-6m^2-33m+2=0$

Use https://en.m.wikipedia.org/wiki/Rational_root_theorem or https://en.m.wikipedia.org/wiki/Polynomial_remainder_theorem

to find $m=2,a=mb=?$