Find the roots of the equation $(1+xi)^n+(1-xi)^n=0$

Question

Find the roots of the equation $f(x)=(1+xi)^n+(1-xi)^n=0$.

I'm having problems finding the roots...this is what I've done:

First I expressed $(1+xi)^n$ and $(1-xi)^n$ in trigonometric form and then I simplified:

$f(x)=2(1+x^2)^\frac{n}{2}(cosn\arctan(x))$ now $1+x^2\neq 0$ because $f(\pm i)\neq 0 $, then

$cosn\arctan(x)=0$ which implies $n\arctan(x)=\frac{\pi(4k+1)}{2}$, $k\in \mathbb{Z}$ and then $x=tan \frac{\pi(4k+1)}{2n}$.

My question is what to do with $k$, because it cannot be any integer?

Sorry if I have a mistake, I'm new.

Answer 1

Let $xi\neq 1$ we have $(\frac{1+xi}{1-xi})^n=-1$ know consider $\frac{1+xi}{1-xi}=w$ so that $w^n=-1=\cos\pi+i\sin\pi$ therefore $$w=\cos\frac{2k\pi+\pi}{n}+i\sin\frac{2k\pi+\pi}{n} \ \ \ \text{ for }k=0,1,2.\ldots, n-1$$on the other hand $\frac{1+xi}{1-xi}=w$ which implies $xi=\frac{w-1}{w+1}$ hence we have $$xi=\frac{\cos\frac{2k\pi+\pi}{n}+i\sin\frac{2k\pi+\pi}{n}-1}{\cos\frac{2k\pi+\pi}{n}+i\sin\frac{2k\pi+\pi}{n} +1}\ \ \ \text{ for }k=0,1,2.\ldots, n-1 $$

Answer 2

In exactly the same spirit as user62498, assuming $(1-ix) \neq 0$, the equation write $$\Big(\frac{1+ix}{1-ix}\Big)^n=-1$$ Using the same notations, we then have $$w=\cos\frac{(2k+1)\pi}{n}+i\sin\frac{(2k+1)\pi}{n} $$ and then $$ix=\frac{\cos\frac{(2k+1)\pi}{n}+i\sin\frac{(2k+1)\pi}{n}-1}{\cos\frac{(2k+1)\pi}{n}+i\sin\frac{(2k+1)\pi}{n} +1}$$ Multiplying by the conjugate of the denominator and simplifying, the rhs of the last equation is just $$i \tan \left(\frac{ (2 k+1)\pi}{2 n}\right)$$ and so $$x= \tan \left(\frac{ (2 k+1)\pi}{2 n}\right)\ \ \ \text{ for }k=0,1,2.\ldots, (n-1)$$ But remember the initial condition $(1-ix) \neq 0$.