JEE Main 2020 (Online) 3rd September Morning ShiftQuestion 1
If $\left(\frac{1+i}{1-i}\right)^{m/2}=\left(\frac{1+i}{i-1}\right)^{n/3}=1$, ($m,n\in\mathbf{N}$) then the greatest common divisor of the least values of $m$ and $n$ is ________.
Complex Numbers
Answer$4$
Solution
$\frac{1+i}{1-i}=i$, so $i^{m/2}=1\Rightarrow \frac m2$ is a multiple of $4$; least $m=8$. $\frac{1+i}{i-1}=\frac{(1+i)(-1-i)}{(i-1)(-1-i)}=\frac{-2i}{2}=-i$, so $(-i)^{n/3}=1\Rightarrow\frac n3$ is a multiple of $4$; least $n=12$. $\gcd(8,12)=4$.