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I have been given this question and didn't find a method without expanding the bracket and solving fourth degree equation.

I used also WA and it solved it using bracket expansion.

Any idea ?

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  • $\begingroup$ Do you know something about roots of unit? $\endgroup$ – Jose Paternina Sep 25 '16 at 23:24
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$\bigg(\dfrac{Z}{1+Z}\bigg)^5 = 1 \implies \dfrac{Z}{1+Z} = \exp{\dfrac{2\pi k i}{5}}$, $k=0,1,2,3,4$.

Finally, solving this, $Z = \frac{\exp{\dfrac{2\pi k i}{5}}}{1 - \exp{\dfrac{2\pi k i}{5}}}$, $k = 0,1,2,3,4$.

However, $k=0$ is not a solution, since division by zero occurs. Hence the answers are valid only for $1 \leq k \leq 4$.


This is consistent with the given polynomial : even though $Z^5$ occurs on both sides, the coefficient is $1$ on both sides, hence after cancellation we are searching for the roots of a degree four polynomial : and we have four roots with us.

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    $\begingroup$ One of these $i$ is not a solution! $\endgroup$ – Semiclassical Sep 25 '16 at 23:29
  • $\begingroup$ Ok, it's not a solution. Then why is the procedure I applied wrong? Do I get a division by zero anywhere? Everything is convincing. Where has the fifth root gone? (the roots are distinct, that's my guess). $\endgroup$ – астон вілла олоф мэллбэрг Sep 25 '16 at 23:38
  • $\begingroup$ in fact one of the five equations is $1+\frac{1}{Z}=1$ which is rejected. so there are only four roots. $\endgroup$ – Mohamed Mostafa Sep 25 '16 at 23:45
  • $\begingroup$ If $i=0$, then you'd be dividing by $1-1=0$. So that's not a solution. This is in fact to be expected: both sides have leading term $Z^5$, so shifting all terms to one side yields a quartic polynomial with only four roots. (Also, you should probably use a different letter than $i$ for the index, since its already used as the imaginary $i$.) $\endgroup$ – Semiclassical Sep 25 '16 at 23:48
  • $\begingroup$ Oh, I see. Thank you for pointing that out. Actually, I did not realize it was quartic in the first place. $\endgroup$ – астон вілла олоф мэллбэрг Sep 26 '16 at 0:18
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Hint: If $Z^5=(1+Z)^5$, then $wZ=1+Z$ where $w^5=1$.

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  • $\begingroup$ It's really good answer thanks $\endgroup$ – Mohamed Mostafa Sep 25 '16 at 23:44

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