In the answer of Surb here : How to solve for the complex number $z$?

I don't understand the subtlety. To me it's natural to do $$z^6=(z-1)^6\iff \left(\frac{z}{z-1}\right)^6\iff \frac{z}{z-1}=e^{ki\pi/3}$$ for $k=0,...,5$. I can see that the case $k=0$, but I don't understand why... We do as usual: $$v^6=1\iff v=e^{ki\pi /3}$$ and now replace $v$ by $\frac{z}{z-1}$, we should get the result. So why it doesn't work here ? Since, by the way, $\frac{z}{z-1}$ is well defined for $z\neq 1$ and that $z=1$ is not solution of $z^6=(z-1)^6$.

  • $\begingroup$ It's very hard to understand what you're missing. You present the calculation, divided to really small, correct, elementary pieces, but you still don't seem to be satisfied. What "doesn't work"? Or can't you solve the equation $z/(z-1) = K$ for a constant $K$? This is a linear equation $z=Kz-K$ i.e. after you multiply it by $z-1$ which is solved by $z=-K/(1-K)$. $\endgroup$ – Luboš Motl Jan 28 '16 at 10:27
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    $\begingroup$ Express $z$, not $v$, as a function of $k$. It does work. $\endgroup$ – Yves Daoust Jan 28 '16 at 10:29
  • $\begingroup$ Take Bernard's hint at Superman's question and note that $v^n=1$ has $n$ different complex solutions. $\endgroup$ – Dietrich Burde Jan 28 '16 at 10:31
  • $\begingroup$ @LubošMotl: What I don't understand is that by calculation we get $\frac{z}{z-1}=e^{ki\pi /3}$ for $k=0,...,5$ whereas, the case $k=0$ doesn't work. So do we have to check that for the other $k$ it work ? And why the calculation doesn't valid the "empirical resolution" ? $\endgroup$ – user301068 Jan 28 '16 at 14:27
  • $\begingroup$ Dear @user301068 - the case $k=0$ doesn't produce any solutions because for that value, the equation reduces to $z/(z-1)=1$ which is equivalent to $z=z-1$ and that has no solutions (because it's equivalent to $0=1$), except for (formally) $z=\infty$. So only the values $k=1,2,3,4,5$ produce five solutions. That shouldn't be surprising because your original equation is a 5th order algebraic equation - the term $z^6$ cancels when you expand the polynomials. $\endgroup$ – Luboš Motl Jan 28 '16 at 15:31

There are $6$ possible values for $v$ that you get from the equation $v^6=1$. But to go from a value of $v$ to a value of $z$ which solves the original equation, you have to solve $v=\frac{z}{z-1}$ for $z$, and when $v=1$ there is no such solution. For if $v=1$, then we would have to have $1=\frac{z}{z-1}$, or $z=z-1$, and this is impossible.

  • $\begingroup$ Alternative way to see that there are just 5 solutions: Rewriting the given equation as $z^6-(z-1)^6=0$, and expanding shows that there actually is no term of degree 6, leaving a polynomial equation of degree 5 to solve. Hence we have only 5 solutions (counting multiplicity, if any). $\endgroup$ – P Vanchinathan Jan 28 '16 at 10:40
  • $\begingroup$ Right, but that still doesn't tell you what went wrong with the argument that seemed to give you 6 solutions. $\endgroup$ – Eric Wofsey Jan 28 '16 at 10:40
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    $\begingroup$ Isn't there already a problem in the first step, from $z^6=(z-1)^6$ to $\left(\frac z{z-1}\right)^6=1$, since this is already assuming $z\ne1$. At that point you just have to check $z=1$ separately, at which point it becomes $1^6=0^6$ so it is not a solution. $\endgroup$ – Mario Carneiro Jan 28 '16 at 10:53

The equation can be written as $ z^6 - (z-1)^6 = 0$.

It can be clearly seen from the above equation that it has five roots, since it is a five degree polynomial. So $z = 1$ is not a solution of the equation.

As mentioned in the question, the value of $k$ can be $1,2 \dots 5$. $k=0$ is not a solution for the equation.

Alternatively, when we are changing the equation to $(\frac{z}{z-1})^6$, we are assuming that $z \ne 1$, so it also rules out the solution for $z=1$.


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