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

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The discussion revolves around finding complex solutions for the equation (x-x^2)(1-x+x^2)^2=1/7. A substitution z = x - x^2 simplifies the problem to solving a cubic equation, leading to further substitutions that ultimately yield a quadratic equation for x. The solutions reveal that all complex numbers x share the same real part of 1/2, while their imaginary parts vary based on specific angles. Comparisons with another solution highlight discrepancies in the real parts, confirming the accuracy of the derived values. The thread concludes with acknowledgments for collaborative problem-solving.
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Find all complex numbers $x$ for which $(x-x^2)(1-x+x^2)^2=\dfrac{1}{7}$.
 
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anemone said:
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.
[/sp]
 
Thanks, Opalg for participating in this challenge! :D

Solution of other:

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 $k=0,\,1,\,\cdots,\,6$.
 
anemone said:
Solution of other:

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 $+\frac12.$
[/sp]
 
Opalg said:
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!
 
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