<s>[INST] 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..
 [/INST] 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 $n\in\Bbb{Z}$.
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