If $\alpha$ and $\beta$ are roots of the equation $x^2-2x+2=0$ then the least value of n for which $\left(\frac{\alpha}{\beta}\right)^n=1$

Question

So simple thought here.

Shoudnt n simply be zero? I mean there is no condition that states n cannot 0, so why is the answer 4? It may be an obvious answer, but i can’t get my head over it.

Thanks!

Answer 1

Let $\dfrac\alpha\beta=r$ we need $$r^n=1$$

$2=(1+r)\beta$

$2= r\beta^2$

$$\dfrac2r=\left(\dfrac2{1+r}\right)^2\iff r^2+2r+1=2r\iff r^2=-1$$

Answer 2

The equation has the solutions $1+i$ and $1-i$. Furthermore $$\frac{1+i}{1-i} = i$$ Although $i^0 = 1$ i think they want $n \in \mathbb{N}, n \geq 1$ what leads you to $n=4$ since $$i^1 = i \neq 1,$$ $$ i^2 = -1 \neq 1,$$ $$i^3 = -i \neq 1$$ $$i^4 =1$$

Answer 3

$x_{1,2} =1\pm i$;

$x_{1,2}= √2e^{\pm i(π/4)}$;

1)$x_1/x_2=i$; 2) $x_2/x_1=-i$;

1)' $(x_1/x_2)^n=i^n=1$ $ \rightarrow$ $n=4,8,..$

2)' Can you finish?