If left(frac{1+i}{1-i}right)^{frac{m}{2}}=lef | vedclass

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(A) First,simplify the base expressions: $\frac{1+i}{1-i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1+2i-1}{2} = i$. $\frac{1+i}{i-1} = \frac{1+i}{-(1-i)} =

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$4$

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(A) First,simplify the base expressions: $\frac{1+i}{1-i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1+2i-1}{2} = i$. $\frac{1+i}{i-1} = \frac{1+i}{-(1-i)} = -\frac{1+i}{1-i} = -i$. Given the equations: $(i)^{m/2} = 1$ and $(-i)^{n/3} = 1$. For $(i)^{m/2} = 1$,the exponent $m/2$ must be a multiple of $4$. Thus,$m/2 = 4k_1 \Rightarrow m = 8k_1$. The least value of $m$ is $8$. For $(-i)^{n/3} = 1$,we know $(-i)^4 = 1$. Thus,$n/3 = 4k_2 \Rightarrow n = 12k_2$. The least value of $n$ is $12$. The greatest common divisor of $8$ and $12$ is $4$.

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

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