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I must to find $z\in\mathbb{C}$ such that:

$\boxed{(1+z)^5=z^5}$

Is the following equivalence correct?

$(1+z)^5=z^5\Leftrightarrow 1+z=z$

If this is not correct, how can solve this problem?

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    $\begingroup$ No, it isn't, because the function $t\mapsto t^5$ is not injetive on $\Bbb C$. $\endgroup$ – user239203 Aug 4 '20 at 4:10
  • $\begingroup$ Thanks Gae, In understood this, thanks. But, how can use this to solve de problem? $\endgroup$ – yemino Aug 4 '20 at 4:13
  • $\begingroup$ math.stackexchange.com/questions/607487/… $\endgroup$ – lab bhattacharjee Aug 4 '20 at 4:36
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Note that $z=0$ is not a solution, so you may divide both sides by $z^5$ and get $$\left(\frac{1+z}z\right)^5=1.$$

Thus for your equation to hold you must have $1+z=z\zeta^r$, where $\zeta=e^{2\pi i/5}$ is a primitive fifth root of unity and $r=0,1,2,3$ or $4$.

Rearranging you get $$z=\frac1{\zeta^{r}-1},$$ for $r=1,2,3,4$. Note there is no solution for $r=0$.

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You have converted a degree $4$ polynomial, which has $4$ roots in $\mathbb C$, into $1 = 0$ which is not true for any $z$. Therefore, this step is definitely invalid.

To find the possible values of $z$, make the substitution $u = z - 0.5$, which transforms the equation into $(u+0.5)^5 = (u-0.5)^5$. Then notice that the terms with even powers cancel, leaving only odd powers. Factoring a $u$ thus gives a quadratic which you can then solve.

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    $\begingroup$ Alternatively, since $z=0$ is not a solution we could divide by sides by $z^5,$ do a substitution $x=1+1/z$ and solve $x^5=0.$ $\endgroup$ – Ragib Zaman Aug 4 '20 at 4:23
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    $\begingroup$ The alternative solution is much better. $\endgroup$ – Robert Israel Aug 4 '20 at 4:28

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