Why is it wrong to take the cube root of both sides $ (z+1)^3=z^3 $

Question

if i take square root of both sides of the equation then the z cancel each other:

$ \sqrt[3] {(z+1)^3 }=\sqrt[3]{z^3} $

$ z+1 =z $

$ z-z+1=0$

$ 1=0 $

I thought this complex equation had no solution but it has if expand the binomial instead of taking the cube root:

$ \sqrt[3] {(z+1)^3} =\sqrt[3]{z^3} $

$ z^3+3z^2+3z+1=z^3 $

$z^3-z^3+3z^2+3z+1=0$

$3z^2+3z+1=0$

which gives us two diferent results:

$-\frac{1}{6}i(\sqrt3-3i)$ and $\frac{1}{6}i(\sqrt3+3i)$

i also checked in wolfram alpha and it seems like i did no mistake. those are the result obtained with each method

1)https://www.wolframalpha.com/input/?i=cuberoot%28%28z-1%29%5E3%29%3Dcuberoot%28z%5E3%29 2) https://www.wolframalpha.com/input/?i=%28z%2B1%29%5E3%3Dz%5E3

In what cases is not correct to take cube root of both sides? i understand that this could be a problem with cube root because the square its not reversible but this is a cuberoot.

Answer 1

You can take cube roots, provided you take into account the fact that you're allowing for complex solutions.

$x^3 = 1$ has three solutions in $\mathbb C$, $x = 1, \omega, \omega^2$. These are called the three complex cube roots of unity. The latter two are non-real conjugate roots, equal to $-\frac 12 \pm i\frac {\sqrt 3} {2}$.

Taking the cube roots, we have $z+1 = z$ which does not give any solutions, or $z+1 = \omega z \implies z = \frac 1{\omega-1}$ or $z+1 = \omega^2 z \implies z = \frac 1{\omega^2-1}$ which will give you the two non-real complex conjugate roots you're looking for.

Alternatively, you could group and factor or expand to solve the resulting quadratic and get the same two complex conjugate roots.

Answer 2

Consider the equation $(x+2)^2=x^2$ instead. Does this imply $x+2=x$ and therefore have no solutions? No: $x=-1$ is a solution.

The problem with just eliminating the $^2$ on both sides is that squaring is not injective: there are different numbers with the same square, so knowing $A^2=B^2$ isn't enough to conclude $A=B$. (For example, we could have $A=1$ and $B=-1$.)

With complex numbers, it's a similar situation. For every integer $n\ge 2$, the operation $z\mapsto z^n$ is non-injective (in fact, every nonzero output value corresponds to $n$ different input values), so you can't safely just eliminate the $^n$ from both sides of an equation.

What you can do is rewrite $A^n=B^n$ as $(A/B)^n=1$ to see that $A/B$ must be one of the $n$ $n$th roots of $1$. That is, $A=\omega^kB$ where $\omega=e^{\frac{2\pi}n i}$ and $k\in\{1,...,n\}$. In this sense, you can "take the $n$th root of both sides" of an equation, but when you do, you have to introduce this additional factor that can take $n$ different values.

Notice that when $n=2$, this gives us $\omega=e^{\pi i}=-1$, so $\omega^k\in\{-1,1\}$, so $A^2=B^2$ is equivalent to $A=\pm B$.

Answer 3

Another way we can interpret the solutions of $ \ (z + 1)^3 \ = \ z^3 \ \ $ is that they are the intersections of the "surfaces" represented by the functions on each side of the equation. While we would recognize immediately that this equation can have no real-valued solutions ("The cube of what number is equal to the cube of one more than that number?"), the surface described by $ \ z^3 \ $ must be considered in $ \ \mathbb{C}^2 \ \ , $ which has a correspondence to $ \ \mathbb{R}^4 \ \ . $ The graphs below (generated by WolframAlpha) show the two three-dimensional "slices" resulting from plotting the real and imaginary parts of $ \ w \ = \ z^3 \ \ $ against the $ \ z \ = \ a + bi \ $ plane. (The three-fold symmetry about the vertical axes can be understood by applying the "polar form" $ \ z^3 \ = \ [ \ \rho · ( \cos \theta + \ i·\sin \theta ) \ ]^3 \ = \ \rho^3 · [ \ \cos(3 \theta) + \ i·\sin(3 \theta) \ ] \ \ . ) \ $ The surface corresponding to $ \ (z+1)^3 \ \ $ would be similar but displaced one unit "to the left" on the $ \ a-$axis. It is the intersections of these complex hyper-surfaces that we seek.

When we treat the equation as $ \ ( \ [a + 1] + bi \ )^3 \ = \ (a + bi)^3 \ \ , $ we obtain $$ (a + 1)^3 \ + \ 3·(a + 1)^2·bi \ + \ 3·(a + 1)·(-b^2) \ + \ (-b^3i) $$ $$ = \ \ a^3 \ + \ 3·a^2·bi \ + \ 3·a·(-b^2) \ + \ (-b^3i) \ \ . $$ Equating the real and imaginary parts of these expressions produces $$ a^3 \ + \ 3a^2 \ + \ 3a \ + \ 1 \ - \ 3ab^2 \ - \ 3b^2 \ \ = \ \ a^3 \ - \ 3ab^2 \ \ \rightarrow \ \ 3a^2 \ - \ 3b^2 \ + \ 3a \ + \ 1 \ \ = \ \ 0 \ \ , $$ $$ 3a^2b \ + \ 6ab \ + \ 3b \ - \ b^3 \ \ = \ \ 3a^2b \ - \ b^3 \ \ \rightarrow \ \ 6ab \ + \ 3b \ \ = \ \ b \ · \ (6a + 3) \ \ = \ \ 0 \ \ . $$ The first of these equations can be understood as that of the "vertical" hyperbola $$ \frac{b^2}{1/12} \ - \ \frac{\left(a + \frac12 \right)^2}{1/12} \ \ = \ \ 1 \ \ , $$ with its center at $ \ a \ = \ -\frac12 \ , \ b \ = \ 0 \ \ $ and its vertices at $ \ \ a \ = \ -\frac12 \ , \ b \ = \ \pm \ \frac{1}{\sqrt{12}} \ \ . $ The second equation represents the union of the "horizontal" line $ \ b \ = \ 0 \ $ and the "vertical" line $ \ a \ = \ -\frac12 \ \ . $ The intersections of these "curves" [as seen in the graph below] are then $ \ z \ = \ -\frac12 \ \pm \ i·\frac{1}{\sqrt{12}} \ = \ -\frac12 \ \pm \ i·\frac{\sqrt3}{6} \ \ , $ exactly the two roots that you found. There are no intersections on the $ \ \mathfrak{Re}(z)-$line, indicating that our equation has no real-valued solutions.

One might well ask "Wait a minute! The equation has third-powers of $ \ z \ \ : $ where is its third root?" The equation $ \ 3z^2 + 3z + 1 \ = \ 0 \ $ that you produced shows that the original equation is actually a quadratic equation "in disguise": the impossibility of the equation being satisfied by any real-number can be thought of as "suppressing" the third root that the cubic equation would have.