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

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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given that,$\left(\frac{1+i}{1-i}ight)^{m} = 1$. First,simplify the expression inside the bracket by multiplying the numerator and denominator by the conjugate of the denominator $(1+i)$: $\frac{1+i}{1-i} 	imes \frac{1+i}{1+i} = \frac{(1+i)^{2}}{1^{2}-i^{2}} = \frac{1+i^{2}+2i}{1-(-1)} = \frac{1-1+2i}{2} = \frac{2i}{2} = i$. Now,the equation becomes $i^{m} = 1$. We know that the powers of $i$ follow a cycle: $i^{1} = i$ $i^{2} = -1$ $i^{3} = -i$ $i^{4} = 1$ 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 the least positive integral value of $m$ is

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