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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

If $x, y \in R$ and $(x + iy)(3 + 2i) = 1 + i$,then $(x, y)$ is

If one root of the quadratic equation $ix^2 - 2(i + 1)x + (2 - i) = 0$ is $2 - i$,then the other root is

If $(1 - i)x + (1 + i)y = 1 - 3i$,then $(x, y) = $

If $x + \sqrt{x^2 + 1} = a$,then $x =$

If $i = \sqrt{-1}$ and $n$ is a positive integer, then $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module