How can $~(z+1)^6 = z^6~$ , where $~z~$ is a complex number?

Question

I don't understand what it is about complex numbers that allows this to be true.

I mean, if I root both sides I end up with Z=Z+1. Why is this a bad first step when dealing with complex numbers?

Answer 1

if so then it would be $$z^6+6z^5+15z^4+20z^3+15z^2+6z+1=z^6$$ or factorized $$ \left( 2\,z+1 \right) \left( {z}^{2}+z+1 \right) \left( 3\,{z}^{2}+ 3\,z+1 \right) =0$$

Answer 2

From

$$(z+1)^6=z^6$$ you draw (taking the sixth root)

$$z+1=\omega^k z$$ where $\omega$ is a primitive sixth root of unity, and $k=0,1,\cdots 5$.

Then for $k\ne0$*,

$$z=\frac1{\omega^k-1}$$ are the five solutions. With $\omega^k=-1$, the solution is real, $z=-\dfrac12$.

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*With $k=0$, $z+1=z$ is impossible.

Answer 3

Try to understand it with real numbers first. $$ y^2=x^2 $$ does NOT mean $$ y=x $$

It means $$ y=\pm x $$ instead. With that in mind, try to understand that a complex circle has even more possible values than just + or minus and thus, such a equation can have a solution. In fact, there is a fundamental theorem that states that in the complex space, EVERY polynomial of degree $n$ has exactly $n$ roots.

Answer 4

Because it is not true (even in $\mathbb R$) that $a^6=b^6\implies a=b$, which is what you are assuming.

In $\mathbb C$,$$a^6=b^6\iff a=b\times\left(\cos\theta+i\sin\theta\right),$$for some $\theta\in\left\{0,\frac\pi3,\frac{2\pi}3,\pi\frac{4\pi}3,\frac{5\pi}3\right\}$.

Answer 5

$$(z+1)^6-z^6=((z+1)^2-z^2)((z+1)^4+(z+1)^2z^2+z^4)=$$ $$=(2z+1)((z+1)^4+2(z+1)^2z^2+z^4-(z+1)^2z^2)=$$ $$=(2z+1)((z+1)^2+z^2-(z+1)z)((z+1)^2+z^2+(z+1)z)=$$ $$=(2z+1)(z^2+z+1)(3z^2+3z+1).$$ Id est, we need to solve $$(2z+1)(z^2+z+1)(3z^2+3z+1)=0.$$ Can you end it now?

Answer 6

Even in $\mathbb{R}$, there is still a solution, namely $z = -\frac{1}{2}$. It is then clear that $(-\frac{1}{2} + 1)^6 = (\frac{1}{2})^6 = (-\frac{1}{2})^6$. In fact, for any 2 real numbers $x$ and $y$ and any positive integer $n$, $x^n = y^n$ if and only if either $x = y$, or $x = -y$ and $n$ is even.