general Complex solution of irreducible quartic polynomial

Question

I encountered a question asking to find the general solution of $x^4+(x-1)^4=0$, surely it has no real solution. I tried $x=(±\sqrt{i}(x-1))$ and $±(\sqrt{-i}(x-1))$, but I was able to find some particular solution only in the form $$\frac{1}{2}+\frac{i\cot(θ/8)}{2}$$ where I got $π/8,7π/8,-π/8,3π/8$ but unable to get the general solution. In the book the solution $$\frac{1}{2}+\frac{i\cot(2k+1)π/8}{2}$$ where $k$ runs from $0$ to $8$. Can anyone please help me in finding the correct form. I am getting $3-4k$ instead of $2k+1$ which is wrong. Also why the book writes $0$ to $8$ for $k$ although its quartic.

Answer 1

Let's go through this slower: clearly $0$ is not a solution, so we can divide by $x^4$: $$1+\left(\frac{x-1}x\right)^4=0$$ $$\left(\frac{x-1}x\right)^4=-1$$ $$1-\frac1x=e^{ik\pi/4}$$ where $k\in\{\pm1,\pm3\}$. Finally $$x=\frac1{1-e^{ik\pi/4}}$$ which, as you have noted, is equal to $$x=\frac12\left(1+i\cot\frac{k\pi}8\right)$$