0

I want to find all complex solutions to $$(z+1)^5 = z$$ My hope was to write it in the form $\alpha^k = 1$ for some $\alpha\in\mathbb{C}$ and $k\in\mathbb{N}$, then calculate the $k$th root of unity. In this case, since $z\neq 0$ that reduces this to $$\biggr(\cfrac{z+1}{z^{1/5}}\biggr)^5 = 1$$ I set $z = re^{i\theta}$. With some algebra the best I could do was that $$(r^{4/5}\exp(4i\theta/5) + r^{-1/5}\exp(-i\theta/5))^5 = 1$$ Now here I'm stuck.

The Art Of Repetition
  • 2,428
  • 1
  • 7
  • 34
Lex_i
  • 2,102
  • 3
    “Most” quintic equations can not be solved exactly in terms of algebraic expressions, and I suspect this is one of them. – Martin R Mar 04 '24 at 09:14
  • Wolfram gives solutions in terms of ${}_4F_3$: https://i.sstatic.net/f7g1W.png, https://www.wolframalpha.com/input?i=solve+z%5E5+%3D+z-1 – Calvin Khor Mar 04 '24 at 09:18
  • 3
    substituting $x=z+1$, your equation is $x^5-x+1=0$. There are two similar famous examples $x^5-x-1=0$ which is non solvable and $x^5+x-1=0$ which is solvable. You could look into this to compute the Galois group – LPZ Mar 04 '24 at 11:05
  • 1
    @LPZ Of course $x^5-x+1=0$ fells into the first category, since replacing $x$ by $-x$ yields $x^5-x-1=0$. :) – Jianing Song Mar 04 '24 at 16:29
  • @JianingSong Doh! Got mixed up in the signs, thought it became something else... Thanks, so the original is unsolvable by radicals. – LPZ Mar 04 '24 at 16:32
  • See Bring radical – Тyma Gaidash Mar 09 '24 at 19:36

1 Answers1

1

Consider the following equation $$x\bigg( 1 + \frac{x}{r} \bigg)^r = x\cdot e_r(x)= z$$ assuming $r \in Q$ and $z \in C$. The solution $x$ is the so-called Lambert-Tsallis function, $W_r(z)$, proposed by R. V. Ramos (please see some answers in my profile). Also, $[e_r(x)]^\alpha = e_{r\alpha}(x\alpha)$. As can be seen, $W_r(z)$ is a multivalued function and it has a lot of interesting properties.

Now, one can write your problem like

$$z^{-1}\bigg(1 +z\bigg)^5 = 1$$ $$z^{-1} \cdot e_5 \bigg(5z\bigg) = 1$$ lets raise to $\color{red}{r=-1}$ $$z^{-r}e_{5r} \bigg(5rz\bigg) = 1$$ which produces $$\frac{W_{5r}(5r)}{5r} = z^{-r}$$

raising again to ^$\frac{1}{r}$ we have

$$ z^{-1} = \frac{1}{\bigg(1 + \frac{W_{5r}(x)}{5r}\bigg)^5 } $$

with $x=5r$

Results After the substitutions $\color{red}{z = \bigg(1 - \frac{W_{-5}(-5)}{5}\bigg)^5 }$

$W_{-5}(-5) = \{ 10.8365 + 0.0000i; 5.9062 - 5.4198i; 5.9062 + 5.4198i; 1.1756 - 1.7624i; 1.1756 + 1.7624i \} $

which produces

$z = \{ -2.1673 + 0.0000i; -1.1812 + 1.0840i; -1.1812 - 1.0840i; -0.2351 + 0.3525i; -0.2351 - 0.3525i\} $

ZKZ
  • 582