Vtyjkyuk

Find all $z$ for the equation
$$(z+1)^7 = z^7$$

The different solutions can be unsimplified and both rectangular or exponential. I have the lead that I subsitute the $z+1$ term with a root of unity. Then, I got lost when I looked at that $z^7$. A thorough explanation would be great!

With $z=w-dfrac12$ then from
$$left(w+dfrac12right)^7=left(w-dfrac12right)^7$$
we have
$$left|w+dfrac12right|=left|w-dfrac12right| ~~~~~,~~~~~ argleft(w+dfrac12right)^7=argleft(w-dfrac12right)^7+2npi$$
the first shows that all points $w$ have the same distance from $dfrac12$ and $-dfrac12$, they are $w=ki$. The second says
$$arctandfrackfrac12=arctandfrack-frac12+dfrac2npi7$$
or
$$arctan2k=dfracnpi7$$
which gives $k=dfrac12tandfracnpi7$, then
$$z=dfrac12+dfrac12tandfracnpi7i~~~,~~~n=0,1,cdots,6$$

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
$$left(fracz+1zright)^7=1.$$

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

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 $left(fracz+1zright)^7 = 1$.

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

Let $psi$ be a (complex) seventh root of unity. Then, the set of solutions is $ frac 11 - 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, $e^frac2pi 17 = psi$ can be taken as a seventh root of unity.