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

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Question

(A) Given the equation: ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^x} = 1$. First,simplify the term inside the bracket by multiplying the numerator an

Vedclass · Accepted Answer

$x = 4n$,where $n$ is any positive integer

Vedclass · Answer

(A) Given the equation: ${\left( {\frac{{1 + i}}{{1 - i}}} ight)^x} = 1$. First,simplify the term inside the bracket by multiplying the numerator and denominator by the conjugate of the denominator $(1 + i)$: $\frac{{1 + i}}{{1 - i}} 	imes \frac{{1 + i}}{{1 + i}} = \frac{{(1 + i)^2}}{{1^2 - i^2}} = \frac{{1 + i^2 + 2i}}{{1 - (-1)}} = \frac{{1 - 1 + 2i}}{2} = \frac{{2i}}{2} = i$. Substituting this back into the equation,we get: $i^x = 1$. We know that $i^k = 1$ if and only if $k$ is a multiple of $4$. Therefore,$x = 4n$,where $n$ is any positive integer.

If ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^x} = 1$,then:

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