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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!

user588857
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1 Answers1

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

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

$0=1$

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 $1/z=0$

Deepesh Meena
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  • Great! How did you get $$1+\frac1z=e^{\frac{2\pi i n}{5}}?$$ – user588857 Sep 07 '18 at 21:33
  • 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. – Ixion Sep 07 '18 at 21:34
  • @Ixion if it isn't $e^{\frac{2\pi i n}{5}},$ then how can it be represented, and why? – user588857 Sep 07 '18 at 21:36
  • $1+\frac{1}{z}=e^{\frac{2\pi i n}{7}}$ with $n=0,1, 2, 3, 4, 5, 6$, because $e^{\frac{2\pi i n}{7}}$ are the 7-rooth of 1. – Ixion Sep 07 '18 at 21:37
  • edited the answer – Deepesh Meena Sep 07 '18 at 21:41
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    @DeepeshMeena Careful with $n=0$ there. For one thing, OP's equation has degree $6$, so it cannot have $7$ roots. – dxiv Sep 07 '18 at 21:45
  • @user588857: Using $e^{2\pi i}=1$, we can see that the 7 distinct numbers $1,,e^{\frac172\pi i},\dots,,e^{\frac672\pi i}$ are all roots of $z^7=1$, which can have at most 7 roots, so these are exactly its roots. – Berci Sep 07 '18 at 21:45