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The method goes like this :

$\displaystyle\frac{(z+2)^{12}}{z^{12}} = 1 $

Let $\displaystyle w = \frac{z+2}z$

$w^{12} = 1$

The solutions to this equation are the $12$ roots of unity.

But one solution, i.e $w=1$, does not work for the original equation. How is it then that we can be sure that the remaining $11$ values of w will work? How do we know that this method works at all? I'd like it if someone could add some rigor to this since it feels very 'flimsy' to just reject one solution by observation, and then not bother to observe the rest.

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  • $\begingroup$ w=1 would imply (z+2)/z = 1 $\endgroup$ – user440261 May 28 '17 at 7:53
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First of all, you are expected to find only $11$ solutions, anyway. The original polynomial has degree $11$. $z^{12}$ can be subtracted on both sides.

In general, it is not unusual to perform operations that temporarily increase the number of solution candidates. The "wrong solutions" will be filtered out later during verification of the result or (in your case) during backward substitution.

In your example, the substitution $w=\frac{z+2}{z}$ introduced another solution candidate (in addition to the $11$ valid solutions), because it is not obvious from the beginning, that $w=1$ won't work.

As long as you only perform operations that do not decrease the number of solution candidates, everything is fine. We do not want to "lose" any solutions. But increasing the number of solution candidates is ok, given that you are going to verify the solutions. This happens often e.g. in case of equations with square roots, in which case you have to perform a lot of squaring to get the candidates.

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  • $\begingroup$ So, because we get 12 solutions, and the polynomial is of degree 11, we discard w=1 since it doesn't work, and we're left with 11 roots of unity. How do we know that there isn't another 'false' root in the solution set of w, and the correct root has to be found by a different method? Basically, how do we know that all the solutions to (z+2)^12=z^12 are included in the solution set of w^12=1? $\endgroup$ – user440261 May 28 '17 at 8:39
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    $\begingroup$ If you do not want to verify the resulting values of $z$ directly, you can argue that $f: \mathbb{C}\setminus\{0\} \rightarrow\mathbb{C}\setminus\{1\} ,\;\; z\mapsto \frac{z+2}{z}$ is a bijection. As we know that $0$ is not a solution of $(z+2)^{12}=z^{12}$, we can exclude $z=0$ from consideration and $(z+2)^{12}=z^{12}$ becomes equivalent with $\left(\frac{z+2}{z}\right)^{12}=1$. The one-to-one mapping between $z$ and $w$ ensures that each solution of $(z+2)^{12}=z^{12}$ in $\mathbb{C}\setminus\{0\}$ matches exactly one solution of $w^{12}=1$ in $\mathbb{C}\setminus\{1\}$ $\endgroup$ – Reinhard Meier May 28 '17 at 9:00
  • $\begingroup$ Thank you. This was just what I was looking for. $\endgroup$ – user440261 May 28 '17 at 9:05
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If we expand $(z+2)^{12}$ by the binomial theorem, the equation is reducible to a polynomial equation of degree $11$. So it has $11$ roots. As OP mentioned, $w=1$ will imply that $z+2=z$ and should be rejected. For the other $11$ roots of unity, we just have no reason to reject them. We can substitute the other values of $z$ found back to the equation and check that they are indeed roots.

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