# Solving the complex equation $(1+z)^5=z^5$ [duplicate]

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?

• No, it isn't, because the function $t\mapsto t^5$ is not injetive on $\Bbb C$. – user239203 Aug 4 '20 at 4:10
• Thanks Gae, In understood this, thanks. But, how can use this to solve de problem? – yemino Aug 4 '20 at 4:13
• math.stackexchange.com/questions/607487/… – lab bhattacharjee Aug 4 '20 at 4:36

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 $$r=0$$.
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.
• 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 $x^5=0.$ – Ragib Zaman Aug 4 '20 at 4:23