Find solution of equation $(z+1)^5=z^5$

Question

I attempt to solve the equation

$(z+1)^5=z^5$.

My first approach is to expand the left hand side but ı get more complicated equation. So I couldn't go further. Secondly, I write equation as, since $z\neq0$,

$(\frac{z+1}{z})^5=1$, put $\xi=\frac{z+1}{z}$

and attempt to solve equivalent equation $\xi^5=1$. But this time it requires more computation to find solutions $z$. Can anyone suggest a simple way to solve this equation? Thanks in advance..

The second approach seems fine. You have $\xi = e^{2\pi i k / 5}$ for $k = 0,\dots,4$. — J. J., Sep 12 '14 at 13:27
Note that you have an easy expression for $1+\frac{1}{z}$. Subtract $1$, take reciprocal. — André Nicolas, Sep 12 '14 at 13:28
It may be useful to think about what kind of mapping $\xi=1+z^{-1}$ is, since once you've solved in $\xi$ you have to deduce which $z$ would have produced these solutions. — Semiclassical, Sep 12 '14 at 13:37

score 3 · Answer 1 · answered Sep 12 '14 at 13:41

3

$(z+1)^5=z^5$ implies that $|z+1|^2=|z|^2$, which implies that $x=\mathrm{Re}(z)=-\frac12$. Then $\left(\frac12+iy\right)^5=\left(-\frac12+iy\right)^5$ is a quadratic equation in $y^2$.

answered Sep 12 '14 at 13:41

Jonas Meyer

55,715

barak manos · Answer 2 · 2014-12-22T15:36:28.483

2

In order to solve $\sqrt[n]{1}$:

Draw the unit circle
Draw the first solution, which is obviously $1+0i=\cos(0)+\sin(0)i$
Repeat $n-1$ times: find the next solution by rotating the previous solution $\frac{2\pi}{n}$ radians

For example, $\sqrt[5]{1}$:

$\cos(0)+\sin(0)i$
$\cos(\frac{2\pi}{5})+\sin(\frac{2\pi}{5})i$
$\cos(\frac{4\pi}{5})+\sin(\frac{4\pi}{5})i$
$\cos(\frac{6\pi}{5})+\sin(\frac{6\pi}{5})i$
$\cos(\frac{8\pi}{5})+\sin(\frac{8\pi}{5})i$

enter image description here

edited Dec 22 '14 at 15:36

answered Sep 12 '14 at 13:51

barak manos

43,599

I suggest using radians as the argument of the trigonometric functions. – Ritz Sep 12 '14 at 13:55
@Ritz: Yes, it has occurred to me while I was writing the answer. Will update; thanks for the advice. – barak manos Sep 12 '14 at 13:55

Hypergeometricx · Answer 3 · 2014-09-12T17:19:32.390

To exploit symmetry, put $z=w-\frac 12$, which gives $z+1=w+\frac 12$.

Solving:

$$\begin{align}(z+1)^5&=z^5\\ \left(w+\frac 12\right)^5&=\left(w-\frac 12\right)^5\\ 2\left[5w^4\left(\frac12\right)+10w^2\left(\frac 12\right)^3+\left(\frac 12\right)^5\right]&=0\\ 80w^4+40w^2+1&=0\\ w^2&=\frac{-40\pm\sqrt{1600-320}}{160}\\ &=-\frac14 \left(1+\frac25\sqrt5\right)\\ w&=\pm\frac i2\sqrt{1+\frac25\sqrt5}\\ z&=-\frac12 \pm\frac i2\sqrt{1+\frac25\sqrt5} \end{align}$$

Alternative method using comments on the OQ:

$$(z+1)^5=z^5\\ \left(\frac{z+1}z\right)^5=1\\ \left(1+\frac 1z\right)^5=1\\ 1+\frac 1z=e^{i2n\pi/5}\\ z=\frac 1{e^{i2n\pi/5}-1} $$

where $n\in\Bbb{Z}$.

Find solution of equation $(z+1)^5=z^5$

3 Answers3

Linked