Find solution of equation $(z+1)^5=z^5$ 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..
 A: $(z+1)^5=z^5$ 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 $y^2$.
A: 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, $\sqrt[5]{1}$:


*

*$\cos(0)+\sin(0)i$

*$\cos(\frac{2\pi}{5})+\sin(\frac{2\pi}{5})i$

*$\cos(\frac{4\pi}{5})+\sin(\frac{4\pi}{5})i$

*$\cos(\frac{6\pi}{5})+\sin(\frac{6\pi}{5})i$

*$\cos(\frac{8\pi}{5})+\sin(\frac{8\pi}{5})i$



A: 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 $n\in\Bbb{Z}$.
