If left(frac{1+i}{1-i}right)^{m}=1,then find | vedclass

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(D) Given $\left(\frac{1+i}{1-i}\right)^{m}=1$. First,simplify the expression inside the parenthesis by multiplying the numerator and denominator by t

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$4$

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(D) Given $\left(\frac{1+i}{1-i}ight)^{m}=1$. First,simplify the expression inside the parenthesis by multiplying the numerator and denominator by the conjugate of the denominator: $\frac{1+i}{1-i} 	imes \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1 + i^2 + 2i}{1 - (-1)} = \frac{1 - 1 + 2i}{2} = \frac{2i}{2} = i$. Substituting this back into the equation,we get $i^m = 1$. We know that $i^n = 1$ when $n$ is a multiple of $4$. The powers of $i$ are $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$. Thus,the least positive integral value of $m$ for which $i^m = 1$ is $m = 4$.

If $\left(\frac{1+i}{1-i}\right)^{m}=1$,then find the least positive integral value of $m$.

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