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I was reading the book Seventeen equations have changed the world.
At some point, while the book was talking about complex numbers, I see this equation:

$2+ \sqrt{-121} = (2+ \sqrt{-1})^3$

Even if it's easy to proof the truth of this equivalence (it is enough to develop the two members),
I can't find an easy/good/fast way to obtain straight the identity.

Can you help me? Does there exist a mathematical property that I'm missing?

mattia.b89
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    You could note that the norm of $2+\sqrt{-121}=2+11\sqrt{-1}$ is $2^2+11^2=125=5^3$, which implies that $2+\sqrt{-121}=\alpha^3$ where the norm of $\alpha$ is $5$. – Servaes Apr 04 '19 at 13:10
  • @Servaes But you would also need $\tan\left(\frac{\arctan(11/2)}{3}\right)=\frac{1}{2}$. Is this easy to see? – kccu Apr 04 '19 at 13:15
  • @kccu I have no idea why you would need that. – Servaes Apr 04 '19 at 13:18
  • I am given $2+\sqrt{-121}$ and I want to write it in some other form. I notice it has norm $5^3$, so it is $\alpha^3$ for some $\alpha$ with norm $5$. How do I know $\alpha = 2+i$? I need to know the argument of $2+\sqrt{121}$, and the $\sin$ and $\cos$ of this argument divided by $3$. – kccu Apr 04 '19 at 13:38
  • https://socratic.org/questions/how-do-i-find-the-cube-root-of-a-complex-number – lab bhattacharjee Apr 04 '19 at 13:40
  • Thus, does not exist a simple reverse formula? Or a sum of cubes? In order to get that result $(2+i)^3$, you have to use a trial-and-error method, that's right? – mattia.b89 Apr 04 '19 at 17:25
  • @kccu No, I need to solve $a^2+b^2=5$, which has the clear solution $a=2$, $b=1$, and three others, but comparing signs already shows that this is the only solution. – Servaes Apr 05 '19 at 10:05
  • @mattia.b89 My answer shows a method that is not purely trial and error. And what do you mean by a reverse formula? – Servaes Apr 05 '19 at 10:08
  • formula such as binomials – mattia.b89 Apr 05 '19 at 17:44
  • @Servaes Right, I was missing that we only needed to look for integer solutions. – kccu Apr 06 '19 at 02:04

4 Answers4

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Expanding the product as \begin{eqnarray*} (2+\sqrt{-1})^3&=&2^3+3\times2^2\sqrt{-1}+3\times2\sqrt{-1}^2+\sqrt{-1}^3\\ &=&(2^3-6)+(3\times2^2-1)\sqrt{-1}\\ &=&2+11\sqrt{-1}\\ &=&2+\sqrt{-121}, \end{eqnarray*} is so elementary that I cannot think of any easier way to show it.

But suppose you want to determine the factorization of $2+\sqrt{-121}$, i.e. roughly speaking an expression $$2+\sqrt{-121}=(a_1+b_1\sqrt{-1})(a_2+b_2\sqrt{-1})\cdots(a_n+b_n\sqrt{-1}),$$ where the $a_i$ and $b_i$ are integers. This is the subject of algebraic number theory. I will use a few results without mention, as I don't know your background in the subject. You could note that the norm equals $$N(2+\sqrt{-121})=N(2+11\sqrt{-1})=2^2+11^2=125=5^3,$$ and that $5$ does not divide $2+\sqrt{-121}$. This implies that $$2+11\sqrt{-1}=(a+b\sqrt{-1})^3,$$ where the norm of $a+b\sqrt{-1}$ equals $5$. This means that $$5=N(a+b\sqrt{-1})=a^2+b^2,$$ which has the solutions $$1+2\sqrt{-1},\qquad 1-2\sqrt{-1},\qquad 2+\sqrt{-1},\qquad 2-\sqrt{-1}.$$ It is then a quick check to show that only $2+\sqrt{-1}$ works.

Servaes
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Not sure how you approached it before.

If you want to utlise the binomial expansion $$ (x+y)^n = \sum_{k=0}^n\left(\matrix{n\\k}\right)x^ky^{n-k} $$ then we have $$ \begin{align} (x+y)^3 &= \left(\matrix{3\\0}\right)2^3 +\left(\matrix{3\\1}\right)2^2\sqrt{-1} + \left(\matrix{3\\2}\right)2^1(\sqrt{-1})^2 + \left(\matrix{3\\3}\right)(\sqrt{-1})^3 \\ &= 2^3 + 3\cdot 2^2\sqrt{-1} + 3\cdot 2\cdot-1 + \sqrt{-1}\cdot(\sqrt{-1})^2\\ &=8+12\sqrt{-1} -6 -\sqrt{-1} = 2 +11\sqrt{-1} = 2 + \sqrt{11^2\cdot - 1} \\ &=2+\sqrt{-121} \end{align} $$

Chinny84
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The development is easy.

Mentally,

$$(2+i)^3=8-6+i(12-1).$$

I can't see any shortcut. (Of course, the identity in the book was established from the right.)

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If it is to prove: $$2+11i=2+i+10i=2+i+(2+i)(2+4i)=(2+i)(3+4i)=(2+i)(2+i+1+3i)=(2+i)(2+i+(2+i)(1+i))=(2+i)(2+i)(2+i).$$

farruhota
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