How do you solve $w^4=16(1-w)^4$?
Giving you answer in Cartesian form.
$\dfrac{w^4}{(1-w)^4}=16$ Are you supposed to let $w=x+yi$?
$\endgroup$ 2$w^4=x^4+4x^3yi-6x^2y^2-4xiy^3+y^4=16$
I then know that you get routes 2,-2,2i,-2i But I dont know how?
I have done $x^4+y^4-6x^2y^2=16$
and $4x^3yi-4xy^3i=0$ => $x^3y=xy^3$ (which doesn't make sense)
Is this the wrong method?
5 Answers
$\begingroup$Let $z=\dfrac{w}{1-w}$. You get $z^4=16$. Note that $2 ^4=16$. Then you need $z'^4=1$. Solve for $z'$, then roll back.
$\endgroup$ 1 $\begingroup$Right away, you should see that $w=0$ doesn't work, so observing that $$16(1-w)^4=2^4(1-w)^4=(2-2w)^4,$$ our problem becomes equivalent to $$1=\frac{16(1-w)^4}{w^4}=\frac{(2-2w)^4}{w^4}=\left(\frac{2-2w}w\right)^4=\left(\frac2w-2\right)^4,$$ meaning we need $$\frac2w-2\in\{\pm1,\pm i\}.$$ Finding those $4$ values of $w$ should then be relatively simple.
$\endgroup$ 2 $\begingroup$If $w^{4}=(2(1-w))^{4}$ you have $w^{2}=(2(1-w))^{2}$ or $w^{2}=-(2(1-w))^{2}$. Solve this two quadratic equations and you should get 4 solutions.
$\endgroup$ 1 $\begingroup$Take a 4th root of both sides (of the question in the title) and $w=2i^\epsilon(1-w)$, where $\epsilon$ can be $0,1,2,3$. Then you just have a linear equation.
$\endgroup$ $\begingroup$It is annoying when the question is changed after an answer has been submitted. Oh well.
Look at the other answers.
I am answering the question in the body of the text, not in the title.
From $\frac{w^4}{1-w^4} = 16$, $w^4 = 16(1-w^4) = 16 - 16 w^4$ or $17 w^4 = 16$ or $w = \sqrt[4]{\frac{16}{17}} =2 \sqrt[4]{\frac{1}{17}} $.
Now use all the 4-th roots of 1 to get the 4 possible values of $w$.
$\endgroup$ 1
