Solving $(z+1)^5 = z^5$
Solution 1:
How about, let $ u = z-1/2 $. Then in terms of $ u $ you have: $$ \left(u+\frac{1}{2}\right)^5 = \left(u - \frac{1}{2}\right)^5 $$ Upon expansion: $$ 5 u^4 + \frac{5}{2} u^2 + \frac{1}{16} = 0 $$ A quadratic equation in $ u^2 $. Can you get it from here?
Solution 2:
If you divide both sides by $z^5$ (note that $z\ne0$, since for $z=0$ we get $0=1$), you get
$$
\left(1+\frac1z\right)^5=1.
$$
The expression in brackets cannot be $1$, so we are left with the four non-trivial fifth roots of unity:
$$
1+\frac1z=e^{2\pi i k/5},\ \ k=1,2,3,4.
$$
So we get four solutions, namely
$$
z=\frac1{e^{2\pi i k/5}-1},\ \ k=1,2,3,4.
$$