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

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Question

(B) First,simplify the expression inside the parenthesis: $\frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} 	imes \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{1^2

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) First,simplify the expression inside the parenthesis: $\frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} 	imes \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{1^2 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i$ Given that ${\left( \frac{1 + i}{1 - i} ight)^m} = 1$,we substitute the simplified form: $i^m = 1$ We know that the powers of $i$ follow the cycle $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$. Therefore,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 integral value of $m$ is:

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