JEE Main 2019 (Online) 8th April Morning ShiftQuestion 1
If $\alpha$ and $\beta$ be the roots of the equation $x^{2}-2x+2=0$, then the least value of $n$ for which $\left(\frac{\alpha}{\beta}\right)^{n}=1$ is
Complex Numbers
Solution
Roots $1\pm i$, $\frac\alpha\beta=\frac{1+i}{1-i}=i$ (or $-i$); $(\pm i)^n=1$ first at $n=4$.