If $\left(\frac{1+i}{1-i}\right)^{\frac{m}{2}}=\left(\frac{1+i}{i-1}\right)^{\frac{n}{3}}=1,(m, n \i

Question

If $\left(\frac{1+i}{1-i}\right)^{\frac{m}{2}}=\left(\frac{1+i}{i-1}\right)^{\frac{n}{3}}=1,(m, n \in N)$ then the greatest common divisor of the least values of $m$ and $n$ is

Professor PC Thomas Classes · Answer

(4) : We have,\ $\left(\frac{1+i}{1-i}ight)^{m / 2}=\left(\frac{1+i}{i-1}ight)^{n / 3}=1$\ $\Rightarrow\left[\frac{(1+i)^{2}}{2}ight]^{m / 2}=\left[\frac{(1+i)^{2}}{-2}ight]^{n / 3}=1$\ $\Rightarrow \quad(i)^{m / 2}=(-i)^{n / 3}=(i)^{4}$\ $\Rightarrow \quad \frac{m}{2}=4 k_{1}$ and $\frac{n}{3}=4 k_{2} \Rightarrow m=8 k_{1}$\ and $n=12 k_{2}$\ So, least values of $m$ and $n$ are 8 and 12 respectively.\ $	herefore \quad \operatorname{GCD}(8,12)=4$\