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?
Asked
Active
Viewed 497 times
1
-
6No, it isn't, because the function $t\mapsto t^5$ is not injetive on $\Bbb C$. – Aug 04 '20 at 04:10
-
Thanks Gae, In understood this, thanks. But, how can use this to solve de problem? – yemino Aug 04 '20 at 04:13
-
https://math.stackexchange.com/questions/607487/the-roots-of-the-equation-zn-1zn – lab bhattacharjee Aug 04 '20 at 04:36
2 Answers
11
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 $r=0$.
tkf
- 15,315
5
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.
Toby Mak
- 17,073
-
4Alternatively, 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 04 '20 at 04:23
-
3