I'm having a little trouble with this problem -

Find all z such that $z^7=(z+1)^7$.

I've moved $(z+1)$ to the other side to obtain $(\frac {z}{z+1})^7=1,$ but I'm not sure how to carry on. I've tried converting to complex exponential form, but how would I simplify and solve? Thanks!


$$z^7=(z+1)^7$$ $$1=\left(\frac{z+1}{z}\right)^7$$ $$1=\left(1+\frac{1}{z}\right)^7$$

$z=0$ is not the solution of the equation because we obtain in the first equation


Thus $$1+\frac1z=e^{\frac{2\pi i n}{7}},\ \ n=1,2,3,4,5,6$$

$$z=\frac1{e^{\frac{2\pi i n}{7}}-1},\ \ n=1,2,3,4,5,6$$

Here $n\ne0$ because when we put $n=0$ in our solution we get $1+1/z=1$ thus $1/z=0$

  • $\begingroup$ Great! How did you get $$1+\frac1z=e^{\frac{2\pi i n}{5}}?$$ $\endgroup$ – user588857 Sep 7 '18 at 21:33
  • $\begingroup$ Sorry, but the values you found are not solutions of the equation. Why is $e^{\frac{2\pi i n}{5}}$? I think you mean $e^{\frac{2\pi i n}{7}}$ with $n=0,..., 6$ instead. $\endgroup$ – Ixion Sep 7 '18 at 21:34
  • $\begingroup$ @Ixion if it isn't $e^{\frac{2\pi i n}{5}},$ then how can it be represented, and why? $\endgroup$ – user588857 Sep 7 '18 at 21:36
  • $\begingroup$ $1+\frac{1}{z}=e^{\frac{2\pi i n}{7}}$ with $n=0,1, 2, 3, 4, 5, 6$, because $e^{\frac{2\pi i n}{7}}$ are the 7-rooth of 1. $\endgroup$ – Ixion Sep 7 '18 at 21:37
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    $\begingroup$ @DeepeshMeena Careful with $n=0$ there. For one thing, OP's equation has degree $6$, so it cannot have $7$ roots. $\endgroup$ – dxiv Sep 7 '18 at 21:45

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