Find the solutions of the equation $(z-1)^5 - (z+1)^5 = 0$

Question

Find the solutions of the equation (where $z$ is a complex number): $$(z-1)^5 - (z+1)^5 = 0$$

According to my book the solutions are $$0, \pm i\cot(\pi/5), \pm i \cot(2\pi/5)$$

I cannot get the same result and I’m asking for help with this.

The book also gives me an identity (which I already proved) $$\frac{1 + e^{(2ik\pi)/n}}{1-e^{(2ik\pi)/n}}= i\cot(\alpha/2)$$ It’s odd that when I tried to get $z$ depending on which of the two $(z-1)^5$ or $(z+1)^5$ I used to divide, the sign changes: $$ z=\pm \frac{1 + e^{(2ik\pi)/n}}{1 - e^{(2ik\pi)/n}}$$ Note: with $z$ positive and negative I got almost every solution given, except for the $0$, and other solutions taking that $e^{(2ik\pi)/5}$ where $k= 0,1,2,3,4$

Answer 1

Note that $z=0$ is not a solution to the original equation. You should be able to see that the $z^5$ terms cancel, leaving a quartic. You have identified the four roots.

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If you want five solutions there is a "solution at infinity" to the original equation, which can be made to make sense. So, if you write $w=\frac 1z$ then you get $(1-w)^5-(1+w)^5=0$ and this is a proper quintic with $w=0$ as a solution. But I think that belongs to a different context from the one in which this equation has been posed.

Answer 2

We can use that

$$(z-1)^5 - (z+1)^5 = 0 \iff \left(\frac{z-1}{z+1}\right)^5=1$$

that is by $w=\frac{z-1}{z+1}$

$$w^5=1 \implies w=e^{\frac{2ik\pi }{5}}\quad k=0,1,2,3,4$$

then solve for $z$ to obtain

$$z=\frac{1 + e^{(2ik\pi)/5}}{1-e^{(2ik\pi)/5}}$$

and using the given identity you are done.

When we have divided by $z-1$ we are assuming $z\neq1$ and that solution needs to be checked in the original equation.

As noticed by Mark Bennet we expect only $4$ solutions since the original equation is a quartic, therefore the case $k=0$ needs to be excluded.

Answer 3

Have you written the problem correctly? z= 0 is obviously NOT a solution to $(z-1)^5- (z+1)^5= 0$. With z= 0, that is $(-1)^5- 1^5= -1- 1= -2$, not 0. It does satisfy $(z-1)^5+ (z+1)^5= 0$.

The first thing I would do is expand each part: $(z- 1)^5= z^5- 5z^4+ 10z^3- 10z^2+ 5z- 1$ and $(z+ 1)^5= z^5+ 5z^4+ 10z^3+ 10z^2+ 5z+ 1$. Subtracting, $(z- 1)^5- (z+ 1)^5= -10z^4- 20z^2- 2= 0$. Now, divide by -2 and let $y= z^2$ to get the quadratic equation $5y^2+ 10y+ 1= 0$. By quadratic formula, $y= x^2= \frac{-10\pm\sqrt{100- 20}}{10}= \frac{-10\pm 4\sqrt{5}}{10}$.