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!
On
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 $z$.
On
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.
With $z=w-\dfrac12$ then from $$\left(w+\dfrac12\right)^7=\left(w-\dfrac12\right)^7$$ we have $$\left|w+\dfrac12\right|=\left|w-\dfrac12\right| ~~~~~,~~~~~ \arg\left(w+\dfrac12\right)^7=\arg\left(w-\dfrac12\right)^7+2n\pi$$ the first shows that all points $w$ have the same distance from $\dfrac12$ and $-\dfrac12$, they are $w=ki$. The second says $$\arctan\dfrac{k}{\frac12}=\arctan\dfrac{k}{-\frac12}+\dfrac{2n\pi}{7}$$ or $$\arctan2k=\dfrac{n\pi}{7}$$ which gives $k=\dfrac12\tan\dfrac{n\pi}{7}$, then $$z=-\dfrac12+\dfrac12\tan\dfrac{n\pi}{7}i~~~,~~~n=0,1,\cdots,6$$