- #1
anemone
Gold Member
MHB
POTW Director
- 3,883
- 115
Find all complex numbers $x$ for which $(x-x^2)(1-x+x^2)^2=\dfrac{1}{7}$.
[sp]anemone said:Find all complex numbers $x$ for which $(x-x^2)(1-x+x^2)^2=\dfrac{1}{7}$.
Very neat solution!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 $k=0,\,1,\,\cdots,\,6$.
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]
The equation we are trying to solve is (x-x^2)(1-x+x^2)^2 = 1/7.
A complex number is a number that can be written in the form a + bi, where a and b are real numbers and i is the imaginary unit equal to the square root of -1.
Solving for complex x allows us to find all possible solutions to an equation, including those that are not real numbers. This is important in fields such as physics and engineering, where complex numbers are often used in calculations.
To solve for complex x, we can use algebraic methods such as factoring, substitution, or the quadratic formula. We may also need to use properties of complex numbers, such as conjugates, to simplify the equation.
There may be multiple solutions to this equation, some of which may be complex numbers. The solutions can be identified by solving the equation and checking if the results satisfy the original equation. In this case, we may need to use a calculator or computer program to find the exact solutions.