MHB Solving for Complex $x$ in $(x-x^2)(1-x+x^2)^2=\dfrac{1}{7}$

[anemone]

Find all complex numbers $x$ for which $(x-x^2)(1-x+x^2)^2=\dfrac{1}{7}$.

 

[Opalg]

Find all complex numbers $x$ for which $(x-x^2)(1-x+x^2)^2=\dfrac{1}{7}$.

[sp]
Let $z = x - x^2$. Then $z(1-z)^2 = \frac17$, from which $7z^3 - 14z^2 + 7z - 1 = 0.$

Solving that cubic equation by Vieta's method, the first step is to make the substitution $z = w + \frac23$. The equation becomes $7\bigl(w + \frac23\bigr)^3 - 14\bigl(w + \frac23\bigr)^2 + 7\bigl(w + \frac23\bigr) - 1 = 0,$ which simplifies to $w^3 - \dfrac13w - \dfrac{13}{7\cdot3^3} = 0.$

The next step is to make another substitution $w = v + \dfrac9v$, which (again after some simplification) gives $v^6 - \dfrac{13}{7\cdot3^3}v^3 + \dfrac1{3^6} = 0,$ a quadratic equation in $v^3$ with solutions $v^3 = \dfrac{13 \pm 3\sqrt3i}{14\cdot27}.$

The modulus of $v^3$ is given by $|v|^6 = \dfrac{13^2 + 3^3}{14^2\cdot27^2} = \dfrac{196}{196\cdot 3^6} = \dfrac1{3^6}.$ Therefore $|v^3| = \dfrac1{3^3}$ and $|v| = \dfrac13.$

The argument of $v^3$ is given by $\arg(v^3) = \arctan\Bigl(\dfrac{3\sqrt3}{13}\Bigr)$, so that $\arg (v) = \theta$, where $\theta$ takes three possible values $$\theta = \frac13\Bigl(\arctan\Bigl(\dfrac{3\sqrt3}{13}\Bigr) + 2k\pi\Bigr) \qquad (k = 0,1,2).$$

Thus $v = \frac13e^{i\theta}$, from which $w = v + \frac9v = \frac13e^{i\theta} + \frac13e^{-i\theta} = \frac23\cos\theta.$

Then $z = w + \frac23 = \frac23(1 + \cos\theta)$. Finally, $x$ is given by $x - x^2 = z = \frac23(1 + \cos\theta)$, so that $x^2 - x + \frac23(1 + \cos\theta) = 0.$ The solutions of that quadratic equation are $$x = \frac{3 \pm i\sqrt{15 + 24\cos\theta}}6. \qquad (*)$$

With the three values for $\theta$ given above, that gives the six complex solutions of the original equation. Notice that they all have the same real part $\frac12.$ For the imaginary part, the numerical value of $\arctan\Bigl(\dfrac{3\sqrt3}{13}\Bigr)$ is approximately $21.787^\circ$, from which you can calculate the three values for $\cos\theta$ as $0.992$, $-0.605$ and $-0.387$. In each case, $15 + 24\cos\theta >0$ so that the square root in $(*)$ is always real.
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[anemone]

Thanks, Opalg for participating in this challenge! :D

Solution of other:

Spoiler

From the identity $(x+y)^7=x^7+y^7+7xy(x^2+xy+y^2)^2$

We deduce $(1-z)^7=1-z^7-7z(1-z)(1-z+z^2)^2$

Hence, our equation is equivalent to $(1-z)^7=-z^7$, that is, $\left(-\dfrac{1}{z}+1\right)^7=1$

It follows that $-\dfrac{1}{z}+1=\cos \dfrac{2k\pi}{7}+i\sin \dfrac{2k\pi}{7}$ for $k=0,\,1,\,\cdots,\,6$.

This reduces to

$\dfrac{1}{z_k}=1-\cos \dfrac{2k\pi}{7}+i\sin \dfrac{2k\pi}{7}=2\sin^2 \dfrac{2k\pi}{7}-2i\sin \dfrac{2k\pi}{7}\cos \dfrac{2k\pi}{7}$

Therefore,

$z_k=\dfrac{1}{-2isin \dfrac{2k\pi}{7}\left(\cos \dfrac{2k\pi}{7}-\sin \dfrac{2k\pi}{7} \right)}=\dfrac{\cos \dfrac{2k\pi}{7}+i\sin \dfrac{2k\pi}{7}}{-2i\sin \dfrac{2k\pi}{7}}=\dfrac{1}{2}\left(-1+i\cot \dfrac{k\pi}{7}\right)$

for $k=0,\,1,\,\cdots,\,6$.

 

[Opalg]

Solution of other:

Spoiler

From the identity $(x+y)^7=x^7+y^7+7xy(x^2+xy+y^2)^2$

We deduce $(1-z)^7=1-z^7-7z(1-z)(1-z+z^2)^2$

Hence, our equation is equivalent to $(1-z)^7=-z^7$, that is, $\left(-\dfrac{1}{z}+1\right)^7=1$

It follows that $-\dfrac{1}{z}+1=\cos \dfrac{2k\pi}{7}+i\sin \dfrac{2k\pi}{7}$ for $k=0,\,1,\,\cdots,\,6$.

This reduces to

$\dfrac{1}{z_k}=1-\cos \dfrac{2k\pi}{7}+i\sin \dfrac{2k\pi}{7}=2\sin^2 \dfrac{2k\pi}{7}-2i\sin \dfrac{2k\pi}{7}\cos \dfrac{2k\pi}{7}$

Therefore,

$z_k=\dfrac{1}{-2isin \dfrac{2k\pi}{7}\left(\cos \dfrac{2k\pi}{7}-\sin \dfrac{2k\pi}{7} \right)}=\dfrac{\cos \dfrac{2k\pi}{7}+i\sin \dfrac{2k\pi}{7}}{-2i\sin \dfrac{2k\pi}{7}}=\dfrac{1}{2}\left(-1+i\cot \dfrac{k\pi}{7}\right)$

for $k=0,\,1,\,\cdots,\,6$.

Very neat solution!
[sp]Comparing my solution with that of "other", the numerical values that I found for the imaginary parts of the six answers work out as $\pm1.0382607$, $\pm0.3987266$ and $\pm0.1141217$. These are the same (to 7 decimal places) as the values $\frac12\cot\frac{k\pi}7 \ (k=1,2,\ldots,6)$ obtained by "other". But of course the solution giving the exact formula for those values is much superior to a mere numerical approximation.

When it comes to the real parts of the answers, I obtained $+\frac12$ in each case, whereas "other" has $-\frac12$. Checking through the solution of "other", I agree with it (apart from the odd typo) up to the line $-\frac{1}{z}+1=\cos \frac{2k\pi}{7}+i\sin \frac{2k\pi}{7}$. After that it should read $$\frac{1}{z_k}=1-\cos \frac{2k\pi}{7} - i\sin \frac{2k\pi}{7}=2\sin^2 \frac{k\pi}{7} - 2i\sin \frac{k\pi}{7}\cos \frac{k\pi}{7}.$$ Therefore $$z_k=\frac{1}{2\sin \frac{k\pi}{7}\left(\sin \frac{k\pi}{7}- i\cos \frac{k\pi}{7} \right)}=\frac{\sin \frac{k\pi}{7} + i\cos \frac{k\pi}{7}}{2\sin \frac{k\pi}{7}}=\frac{1}{2}\left(1+i\cot \dfrac{k\pi}{7}\right),$$ so that the real part is in fact $+\frac12.$
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[anemone]

Very neat solution!
[sp]Comparing my solution with that of "other", the numerical values that I found for the imaginary parts of the six answers work out as $\pm1.0382607$, $\pm0.3987266$ and $\pm0.1141217$. These are the same (to 7 decimal places) as the values $\frac12\cot\frac{k\pi}7 \ (k=1,2,\ldots,6)$ obtained by "other". But of course the solution giving the exact formula for those values is much superior to a mere numerical approximation.

When it comes to the real parts of the answers, I obtained $+\frac12$ in each case, whereas "other" has $-\frac12$. Checking through the solution of "other", I agree with it (apart from the odd typo) up to the line $-\frac{1}{z}+1=\cos \frac{2k\pi}{7}+i\sin \frac{2k\pi}{7}$. After that it should read $$\frac{1}{z_k}=1-\cos \frac{2k\pi}{7} - i\sin \frac{2k\pi}{7}=2\sin^2 \frac{k\pi}{7} - 2i\sin \frac{k\pi}{7}\cos \frac{k\pi}{7}.$$ Therefore $$z_k=\frac{1}{2\sin \frac{k\pi}{7}\left(\sin \frac{k\pi}{7}- i\cos \frac{k\pi}{7} \right)}=\frac{\sin \frac{k\pi}{7} + i\cos \frac{k\pi}{7}}{2\sin \frac{k\pi}{7}}=\frac{1}{2}\left(1+i\cot \dfrac{k\pi}{7}\right),$$ so that the real part is in fact $+\frac12.$
[/sp]

Ops...yes, that is another careless mistake and to be honest, I saw your $+\frac36$ and thought I made no typo, without realizing the minus sign in front...thank you Opalg for catching that!