How to solve the equation $(z+1)^7 = z^7$ for all $z$?

Solution 1:

You're on the right track with roots of unity. A clue is that both sides have an expression to the same power. Dividing both sides by $z^7$, we get $$\left(\frac{z+1}z\right)^7=1.$$

Next, you could let $u = \frac{z+1}z$, write out the solutions to $u^7=1$, then convert back to $z$.

Solution 2:

Note that $z = 0$ is not a solution to the equation, so anything which satisfies the equation $(z+1)^7 = z^7$ must also satisfy the equation $\left(\frac{z+1}{z}\right)^7 = 1$.

Consequently, $1 + \frac 1z = \frac{z+1}{z}$ must be a seventh root of unity.

Let $\psi$ be a (complex) seventh root of unity. Then, the set of solutions is $\{ \frac 1{1 - \psi^k} : 1 \leq k \leq 6\}$, since $1 + \frac 1z = \psi^k$ for some $1 \leq k \leq 6$. Note that there are six distinct solutions, and your equation can be reduced to the roots of a degree six polynomial, so you can verify the answer like this.

(Note that $\psi^7 = 1$ is not a root of the polynomial, clearly)

Note : In polar form, $e^{\frac{2\pi 1}7} = \psi$ can be taken as a seventh root of unity.