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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!

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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 $$z=-\dfrac12+\dfrac12\tan\dfrac{n\pi}{7}i~~~,~~~n=0,1,\cdots,6$$

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  • $\begingroup$ When $n = 0$, we have $z = 1/2$, but $(3/2)^2 \ne (1/2^2)$, so there are only $6$ solutions, in keeping with the fact that $z^7$ cancels out of $(z +1)^7 = z^7$, leaving the $6$-th degree equation $z^6 + z^5 + \ldots + z + 1 = 0$. $\endgroup$ Sep 19 '18 at 19:05
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    $\begingroup$ What do you suggest in this case? $\endgroup$
    – Nosrati
    Sep 19 '18 at 19:10
  • $\begingroup$ I think all solutions for $n \ne 0$ are likely correct. Let me check . . . Cheers! $\endgroup$ Sep 19 '18 at 19:11
  • $\begingroup$ I would start with the observation that neither $z = 0$ nor $z = -1$ satisfy $(z + 1)^7 = z^7$ $\endgroup$ Sep 19 '18 at 19:14
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    $\begingroup$ Sorry, the mistake it here $$z=w-\dfrac12=z=i\dfrac12\tan\dfrac{n\pi}{7}-\dfrac12$$ thank you for your checking the answer. $\endgroup$
    – Nosrati
    Sep 19 '18 at 19:19
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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 $z$.

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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.

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